LIBRARY OF CONGRESS. 




Shelf JlS^ 

UNITED STATES OF AMERICA. 



■■HHMB 



AN 



Elementary Treatise 



ON 



M ECHAN ICS 



C. J. KEMPER ° 

Professor of Mathematics and Modern Languages, Bethany College, West Va. 



V 



< 
NEW YORK 



PRINTED BY S. W. GREEN'S SON 
74 and 76 Beekman Street 

1882 






Copyright, 1882, 

BY 

C. J. KEMPER 



PREFACE. 



The preparation of the following pages is due in part to 
the interesting nature of the subject itself, and partly to the 
confident belief that it is susceptible of a treatment somewhat 
different from that usually adopted, without departing in any 
degree from the strict methods of mathematical analysis. 
Many of the works on Theoretic Mechanics, by distinguished 
authors, — works of high merit so far as relates to the able treat- 
ment of the subject-matter, — are nevertheless placed beyond 
the reach of large numbers of students on account of the ex- 
clusive use of the methods of the Calculus. There are very 
few students in our colleges who, even after having finished a 
course in the Differential and Integral Calculus, possess any 
very clear apprehension of the abstract truths involved in the 
pure theory, and consequently few are qualified to under- 
stand their more difficult applications to the complicated 
forces of physical nature. 

Other works adopt in some portions the methods of the 
Calculus, and in others the ordinary Algebraic and Geometric 
methods, which detracts from their adaptation to that larger 
class of students unacquainted with the Calculus, as well as 
from the uniformity of the text. One prime object in the 
following treatise has been not only to avoid the Calculus 
entirely, except in one or two instances for illustration, to 
confirm results already reached, but to do so without any 
sacrifice of thoroughness or comprehensiveness. Many of the 
most difficult portions of the subject have been, after very 
considerable labor and thought, brought down to a lower 



IV PREFACE. 

plane of treatment. With this design always in view, works 
adopting the higher analysis have been especially consulted, 
for the purpose of choosing the most interesting material for 
simplification. How far success has crowned the effort to 
place this attractive study within reach of a greater number of 
students will be left for others to judge. Perhaps the attempt 
has met with more signal success in that portion of the work 
devoted to Central Forces than any other. It is believed, how- 
ever, that other portions will be found much simplified, as in 
some parts of the treatment of the Centre of Gravity, the The- 
orem of Pappus, Moment of Inertia, Motion on Inclined Planes, 
the Compound Pendulum, etc. In fine, the difficulties of the 
subject have never been shunned to avoid the labor and appli- 
cation necessary to present them in harmony with the plan 
adopted, which it is to be hoped will be duly appreciated by 
those who may thus have the subject presented in a way com- 
mensurate with their mathematical attainments. 



CONTENTS. 



Sections 
Definitions and Explication of Terms. — Mechanics. — Inertia. — 

Mechanical Force. — Measure of Force. — Mass. — Attraction. — 

Elasticity. — Representation of Forces. — Resultant and Component 

Forces I to 12 

Part First. — Statics. 

Parallelogram of Forces. — Demonstration of the Theorem. — Triangle 
of Forces. — Polygon of Forces. — Method of Leibnitz for determining 
the resultant of concurring forces 12 — 18 

Application of foregoing principles. — Vibration of Pendulum. — Motion 

on Inclined Planes. — Examples for solution 18 — 24 

Composition and Resolution of Forces. — Concurring Forces. — De- 
termination of the Resultant by reference to rectangular axes. — For- 
mulae. — Examples 24 — 26 

Two Concurring Forces in One Plane. — Resultant. — Examples. — 
General proposition of three forces in equilibrium. — Examples for 
solution 26 — 28 

Concurring Forces not in the Same Plane.— Resultant by reference 

to three axes. — Examples for solution 28 — 33 

Non-Concurring Forces.— Parallel Forces in the same plane.— Resul- 
tant. — Examples. — Opposite parallel forces 33—37 

Of Moments. — Definition. — Static Couples. — Propositions relating 
thereto. — Revolution of couple in its own plane. — Transference to 
parallel plane. — Composition of Rotations 37 — 47 

Geometrical Construction of Parallel Forces.— Of Parallel Forces 
not in the same plane. — Resultant. — Co-ordinates of its point of ap- 
plication. — Of moments with respect to a plane. — Conditions of 
equilibrium 47 — 56 

Of Oblique Non-Concurring Forces in the Same Plane. — Resul- 
tant. — Graphical solution. — By reference to rectangular axes. — Of 
moments referred to a point. — Conditions of equilibrium of the sys- 
tem 56 — 63 



VI CONTENTS. 

Sections 
Of Oblique Non-Concurring Forces in Space. — Reference to three 

axes. — Co-ordinates of point of application of resultant. — Of mo- 
ments with respect to an axis. — Conditions of equilibrium of the 
system 63 — 73 

Centre of Gravity. — Examples. — Triangle. — Polygon. — Pyramid. — 
Spherical Zone. — Spherical Sector. — General Method for Centre of 
Gravity. — Examples. — Circular Arc. — Sector. — Segment. — Semi-El- 
lipse. — Portion of a Parabola 73 — 81 

Theorem of Pappus. — Demonstration. — Centrobaryc method of de- 
termining solids and surfaces of revolution. 81 — 82 

Theorem of Leibnitz. — Illustrative problems for solution 82 — 86 

The Mechanical Powers. — Lever. — False Balance. — Wheel and 
Axle. — Differential Axle. — Pulleys. — Inclined Plane. — Wedge. — 
Screw. —Conditions of equilibrium and mechanical advantage of the 
various powers. — Friction. — Coefficient of Friction. — Rolling Fric- 
tion 86 — 108 

Principle of Virtual Velocities. — Demonstration. — Applications. — 

To the bent lever ; the movable pulley ; the inclined plane 108 — no 

Part Second. — Dynamics. 

Of Motion in General. — Rectilinear motion. — Uniformly varied mo- 
tion. — Formula for the motion of a free body solicited by a constant 
force. Application to Force of Gravity. — Acceleration as a relative 
measure of forces no — 122 

Of the Motions of Bodies upon Inclined Planes. — Time of descent 
along chords of a vertical circle. — Velocities acquired, and times of 
descent upon planes however inclined. — Velocities upon curves equal 
to that attained in falling through the height of the curve 122 — 135 

Theory of the Simple Pendulum. — Time of vibration as square root 
of length. — General expression for the time of vibration. — Applica- 
tion to determining the force of gravity at any station 135 — 148 

The Cycloidal Pendulum. — Expression for time of vibration. — Isoch- 
ronous .... 148 — 149 

Central Forces. — This division embraces a full discussion of the laws 
which obtain in the motion of bodies around a fixed centre of force, 
as applicable to Circular Orbits ; Elliptic, Parabolic, and Hyperbolic 
Orbits. — The Logarithmic Spiral, Lemniscala, etc. — Kepler 's Laws. — 
Modification of the third law for the solar system. — Species of the 
orbit as dependent upon initial velocity. — General method of deriving 
the law of the force for any orbit 149 — 1 75 



CONTENTS. Vii 

Sections 
Of Centrifugal Forces. — Pendulum vibrating in a vertical plane. — 

Centrifugal Force upon the earth's surface. — Determination of the 

absolute Force of Gravity and Centrifugal Force. — Their ratio. — 

Centrifugal Forces of extended masses. — Same as if whole mass were 

collected at the Centre of Gravity. — Examples for solution i75_jg 2 

The Conical Pendulum. — Time of vibration. — Constant for cones of 
equal altitude. — Tendency to elliptic motion in small orbits ex- 
plained. Application to curved roadways. — Modification for rail- 
road curves. — Motion upon the inner surface of a Paraboloid of Revo- 
lution o 182—187 

Moment of Inertia. — General formula. — Application to material 
line. — Rectangular Plate. — Circumference of Circle. — Circular 
Plate. — Spherical Shell. — Solid Sphere, etc; 187 — 191 

Centre of Gyration. — Radius of Gyration. — Formula. — Principal 

Radius of Gyration 191 — 192 

Principle of D'Alembert. — Applications. — To movement on In- 
clined Planes. — Atwood's Machine. — Wheel and Axle 192 — 195 

Compound Pendulum. — Equivalent Simple Pendulum. — Centre of Oscil- 
lation. — Solution of Examples. — Simple Rod. — Circular Plate sus- 
pended. — Axes of Oscillation and Suspension reciprocal. — Practical 
method of determining the seconds pendulum. 195 — 200 

Centre of Percussion. — Definition. — How related to Centre of Oscil- 
lation. — Axis of Spontaneous Rotation 200 — 202 

Motion of Centre of Gravity. — When uniform and rectilinear. — 
When the Centre of Gravity remains stationary. — Effect of com- 
bined action of impressed forces when applied to Centre of Gravity. — 
Motion of translation and rotation by the same impulse. — Centre 
of Spontaneous Rotation. — How related to Centre of Oscillation. — 
Examples. — Application to orbital and diurnal motion of the earth. . 202 — 208 

Appendix 208 — 



Elements of Mechanics. 



Definitions and Explication of Terms. 

[I.] Mechanics. — The science of Mechanics treats of the 
laws of equilibrium and motion of matter. With reference to 
solids the subject is divided into Statics and Dynamics, the 
former treating simply of the laws of equilibrium of bodies at 
rest, and the latter of the laws which govern them in motion. 

In the case of fluids the corresponding divisions of the 
subject are embraced in the terms Hydrostatics and Hydro- 
dynamics, treating respectively of the laws of equilibrium and 
motion of fluids ; and to the gases and elastic fluids are applied 
the analogous terms Aerostatics and Aerodynamics. 

Under these heads may be included whatever is properly 
embraced in an elementary treatise on Mechanics. 

[2.] Inertia. — By the Inertia of matter is meant the prin- 
ciple to which is due the resistance which it offers to an im- 
pressed force, tending to change its state, whether it be in 
imparting or destroying motion, reference being had only to 
that resistance to change of condition as to motion, which is 
independent of any molecular attractions, or repulsions, or 
external media. 

The term Inertia is, however, singularly unfortunate, in 
that etymologically it expresses an idea the very opposite of 
that intended to be conveyed, since it is rather the negation 
than the affirmation of the positive attribute of matter which 
it is used to designate. Very little consideration will be 



ELEMENTS OF MECHANICS. 

necessary to convince us that the term Vis Inertia, or force of 
Inertia, is even more inappropriate, as embodying contradic- 
tory ideas. In the present treatise, however, the term wiil be 
used in its technical or conventional sense. 

That such a property really resides in matter, becomes 
known to us only from experience or observation. Nor is it 
difficult to find conclusive illustrations of this truth. 

If we let fall a heavy ball, and immediately increase its 
downward velocity by a rapid pressure of the hand from 
above, the resistance, perceptibly felt, is that chiefly due to the 
Inertia of the mass. The slight increase of resistance, due to 
the increase of velocity through the air as a resisting medium, 
is easily shown to be inappreciable by the rapid motion of the 
extended hand alone through the air. 

Again, if matter possessed within itself no power of resist- 
ing motion, the motion of a body, if all possibility of resistance 
from other causes be removed, would be equal to that of the 
force by which it is solicited. 

Under this supposition, therefore, a body falling in vacuo, 
since, whatever other external forces might act upon it, there 
would be a resultant gravitating force, would move with the 
velocity of the gravitating force itself, in its action. 

But since the force of gravity is ever present and active, 
it is equivalent to a force acting with indefinite velocity, and 
hence the motion of a body in vacuo should be of the same 
character ; whereas experiment shows that this is by no means 
true ; the retardation of the body being due to a developed 
resistance from within. 

It is evident that the impartation of motion to a mass, 
does not bestow upon it any peculiar property to resist an 
acceleration, or retardation, which it did not originally pos- 
sess ; the only new property acquired being that of the 
motion itself. But under this motion the mass per se, or 
uninfluenced by any extraneous force which develops resist- 
ance, is entirely passive whatever may be its velocity ; and if 



DEFINITIONS AND EXPIICATION OF TERMS. 3 

we suppose it to be acted upon by a given force, the accumu- 
lation or change of velocity, will depend not upon the velocity 
already existing, but upon the applied force and the power of 
resisting motion residing in the mass. In other words, the 
property of resistance to change, or the Inertia of a mass, is 
independent of any existing velocity ; but the developed 
resistance to change due to the application of an extraneous 
force, is proportional directly to the mass and the velocity im- 
parted in a given time ; and hence the force required to over- 
come this resistance, or rather to balance it, as being devel- 
oped under motion, is also directly proportional to the mass 
and the velocity imparted. If the impressed force ceases to 
act, the motion becomes uniform ; if it continues to act, it 
balances a newly developed resistance logically dependent 
upon increased velocity. There is a destruction of applied 
energy, as such, but a conservation of the same energy as 
stored up in the moving mass. A simple change of form. 

The Force of Inertia, then, is simply a developed resistance 
to change, consequent upon the application of some extrane- 
ous force, and at any instant of the accumulation or destruc- 
tion of velocity is equal and opposed to the impressed force, 
the change of velocity being a result and necessary condition 
of the development of the Force of Inertia. This force is also 
proportional to the mass, for, being a resistance to change 
exerted by each particle of the mass, the aggregate result 
must depend upon the number of these particles, or the mass 
itself. 

[3.] Mechanical Force.— That which tends either to 
produce or destroy the motion of bodies is termed Force. 
Forces may be of various kinds, all tending, however, to over- 
come the Inertia of matter and other resisting forces ; which 
is the direct effect of their application, whilst motion is the 
ordinary result. A force may act instantaneously, as by a 
single impulse ; or it may act uniformly, as a constant force, 
unchanging in its intensity; or as the force of gravitation, 



4 ELEMENTS OF MECHANICS. 

which decreases as the squares of the distances between the 
masses developing it increase; and although we may con- 
ceive the possibility of a force acting according to almost 
any law, the three cases mentioned, constitute the three forms 
most common to the forces of nature, and should therefore 
be chiefly considered. 

If we conceive a body to be acted upon by single impulses 
at equal intervals of time, and then suppose these intervals to 
become gradually less and less, this kind of force would con- 
tinually approximate to a constantly acting force, and finally, 
should the intervals of time become infinitesimal, would in 
effect constitute such a force. As will be seen hereafter, this 
consideration will be of importance in interpreting the effects 
of such forces. If, moreover, the impulses be supposed equal, 
the resulting force will be constant, or unchanging in intensity. 
Such a force will be designated a constant force. 

[4.] Measure of Force. — If a force F imparts by a single 
impulse to a mass M a velocity V, a double impulse would 
impart to the same mass a double velocity [2] ; and since a 
double velocity would in a given time accomplish a double 
space, therefore either the velocities imparted or the spaces 
accomplished in a given time would become relative measures 
of the two forces. Two constant forces may also be compared 
with each other by means of the velocities or spaces gener- 
ated in a given time, just as two simple impulses. 

To render this apparent, conceive two forces, each of 
which in a time T acts by n equal impulses at equal inter- 
vals of time, and let each impulse of one force be m times the 
measure of a single impulse of the other. 

If v equals the velocity imparted by each of the n equal 
and successive impulses of the lesser force, then nv will rep- 
resent the velocity acquired in the whole time T. 

But since every impulse of the greater force is m times an 
impulse of the lesser, the velocity imparted by the former for 
one impulse would be m X v, and for the whole time T the 
velocity acquired would be m X v X n. 



DEFINITIONS AND EXPLICATION OF TERMS. 5 

But n X v ', m X v X # 11 i : m; 

that is, the velocities imparted in the time T are as the forces. 

In the second place, if s denote the space accomplished 
by the lesser force during the first interval between the im- 
pulses, then during the second interval the space will be 
described with a velocity due to two impulses, or a double 
velocity, and will consequently be 2 s, during the third interval 
with a treble velocity, which will give a space 3 s, and finally, 
for the nth interval, a space n . s, and the whole space for the 
given time will be 

S + 2S+$S + + nS = s(l +2 + $ +«)• 

In like manner, if a second force be m times as great, the 
space for the first interval would be m s, and the whole space 
for n intervals in the time T would be 

m X s (1 + 2 + 3 . . . . + n). 

But these results are to each other as 

s : m . s : : i : m. 

That is to say, the whole spaces passed over in a given time are as 
the forces. 

Now since the above is true, whatever be the number n 
of intervals in the time T, it will be true when n becomes 
Infinite and the intervals of time infinitesimal, in which case, 
however, [3] the two forces become in effect constant forces. 
Hence we conclude that two constant forces acting upon the same 
or equal masses, are likewise to each other as the velocities gener- 
ated, or spaces accomplished in the same time. 

Hence if F and F l denote two forces, either impulsive or 
constant, V and V x the velocities they would impart to a mass 
M in a given time, and 5 and 5 X the spaces accomplished, we 
should have 

F\F X :: V: V x :: S: S,. 

Let us now consider the general case, in which two unequal 



O ELEMENTS OF MECHANICS. 

forces act upon unequal masses to produce rectilinear motion. 

Let F and F 1 be the two forces ; M and M x the two masses, 

composed respectively of unequal numbers of equal particles 

denoted by m and m 1 ; and Fand V 1 their respective velocities. 

If we conceive the force Fto be distributed in its action 

F 
amongst the particles of the mass M, we shall have - for the 

F 
force acting upon each particle. Similarly — — will represent 

the force upon each particle of M x . 

The velocities of the masses would plainly be the same as 
if the forces were not distributed in their action, and would 
be equal to the velocities of each particle in the respective 
masses. 

But these latter velocities, from what has already been 
shown, will be as the forces acting upon each particle, and 
we shall have 



^.5 

m ' m, 



V: V x . 



We have also m : m x : : M ' : M v 

whence, by multiplying the corresponding terms of the pro- 
portions, we have 

F\F X \\ MV\M X V x . 

The product of a mass M by its velocity V is called the 
quantity of motion of the mass. Hence any two forces are to 
each other as the quantities of motion they would respectively 
generate in the same time or in the unit of time. 

The term quantity of motion, an injudicious one, expresses 
therefore the relative measure of a force generating motion, 
in comparison with other like forces. 

Cor. i. — If in the proportion above we make V — V lf we 
have F : F 1 :: M : M x . That is, the velocities being equal, the 
forces will be as the masses. And conversely, if F : F x : : M : M lf 
then V=V X \ or, the forces being as the masses, the velocities will 
be equal. 



DEFINITIONS AND EXPLICATION OF TERMS. *] 

Cor. 2. — If we suppose F — F v we have M V — M x V lf or 

MV 
V x = M . In this equation V 1 is the velocity the force F v or 

its equal F, will impart to M 1 in a given time. Hence we can 
easily find the velocity a force Fwill impart to any mass M v ivhen- 
ever we know the quantity of motion it will generate in a?iy other 
mass M in the same time. 

Cor. 3. — Since the Inertia of a mass, or its power to resist 
change of condition as to motion, is independent of its velo- 
city, it is obvious that in every instant of time a constant force, 
whilst it develops equal resistances, will impart equal incre- 
ments of velocity. Hence under the action of such a force 
the velocities imparted will be proportional to the times. The 
force of gravity for small distances from the earth's surface 
may be regarded as a constant force, and hence if v denote 
the velocity it will impart to a mass in the time /, and g the 
velocity imparted to the same mass in one second, we shall have 

v : g :: t : 1", or v =gt. 

[5.] Mass. — The mass of a body is almost uniformly de- 
fined by writers on the subject as the quantity of matter it 
contains, and the weights of any two bodies at the same lati- 
tude are assumed as the relative measures of their masses, or 
quantities of matter. This of course proceeds upon the hypo- 
thesis that the weight is not affected by any quality of matter 
entirely distinct from and independent of its mass, as far as 
regards the earth's attraction. This is doubtless true ; but as 
its truth could not be determined a priori, it needs to be 
confirmed by experience. Our notion of mass, however, is 
entirely distinct from that of weight. It is sui generis, and 
apparently stands more nearly related to inertia than to weight. 

Since the centrifugal force of a mass, revolving in a hori- 
zontal circle, is due to its inertia, and this latter we conceive 
to be proportional to its quantity of matter, experiment can be 
made use of to practically confirm the theory that weight may 
also be adopted as a relative measure of masses at the same 



8 ELEMENTS OF MECHANICS. 

locality. We prefer, therefore, not to attempt in so many 
words a definition of mass, or quantity of matter; but as it is 
necessary in dynamical science to establish at least a relative 
measure for this new species of quantity, as pertaining to dif- 
ferent bodies, we assume that the mass of a body is proportional 
to its inertia, or inversely to the velocity that a given force acting 
in a definite manner for a given time, will impart to it. 

Assuming also that the mass of any body is proportional 
to its weight, then, since the weight changes with the lati- 
tude, it becomes necessary to derive an expression for the 
mass which shall be independent of the latitude of the place. 

If, then, we denote by g the velocity imparted to a mass m 

in the unit of time, or one second, by gravity, and by J^its 

weight, the quantities g and W will both be directly propor- 

W 
tional to the intensity of this force, and hence the ratio — 

will be a constant quantity for all latitudes, and may there- 
fore be assumed to represent the mass, which is also constant. 
Thence we have 

W 

m = — • 



[6.] The relations deduced in the preceding articles evi- 
dently apply either to impulsive or constant forces. Since a 
constant force imparts an equal velocity to a mass, in every 
instant of time, the motion induced would be unifor7nly acceler- 
ated, and the force is called a uniformly accelerati?ig force. If 
the force should not be constant, but vary in intensity at suc- 
cessive instants of time — as, for example, in the case of gravi- 
tation, where the force increases with the decrease of distance 
between the masses developing it — the acceleration would be 
variable, and the force a variable accelerating force. 

The intensity of a variable force, or force of varying inten- 
sity, can of course only be referred to a given instant of time. 

The proportions of [4] furnish only the relative measures 
by which forces may be compared with each other. To de- 
duce an absolute standard or measure for any force, we must 



DEFINITIONS AND EXPLICATION OF TERMS. 9 

compare it with some determinate force existing in nature — 
as, for example, the force of gravity — or with some other arti- 
ficial standard assumed as a unit force. 

If we suppose F to denote at any instant of time the inten- 
sity of a variable force, we may consider that for an indefi- 
nitely short time t it is of uniform intensity. If, therefore, V x 
denote the velocity it is capable of imparting in the time /, 
and V the velocity imparted by gravity in the same time, we 
shall have [4], representing the force of gravity by W, the 
weight of the body, 

F: W :: V x : V. 
But, as already shown [4, Cor. 3], V = gt; 

V g t t 

That is to say, the intensity of a variable force at any instant of 
time is proportional to, or varies as the product of the mass by the 
velocity generated in an indefinitely small time, divided by the time. 

That a force of varying intensity may for an infinitely 
short time, be considered as invariable, will appear from the 
fact that the variation in the value of the force for the given 
time, will be infinitely small in comparison with the finite 
value of the whole force, and hence may without error be 
neglected. A similar remark will obviously apply to a vary- 
ing velocity. 

If F x denote a second force imparting a velocity F 2 to a 

V 
mass M lf we should in like manner have F x = M x — -, 

and F: F x :: MV X : M\V„ 

the velocities V x and F 2 being such as are imparted in the 
time t, during which interval the forces are supposed to be 
of uniform intensity ; or, any two forces are to each other as the 
quantities of motion they can impress in the same indefinitely 
small time. 



IO ELEMENTS OF MECHANICS. 

[7.] We have seen that the measure of any constant motive 
force is equal to the mass moved multiplied by the velocity 
imparted, or the acceleration, in the unit of time. 

That is, F = M v. If we suppose the force F to be equally 

distributed in its action upon the units of mass contained in 

F 
the whole mass M, then -^> will represent the force acting 

F 
upon the unit mass. But t>= v, is equal to the acceleration 

for the unit of time imparted either to the mass M, by the 

F 
whole force F, or to the unit of mass, by the force -^. But 

the accelerations of the unit mass in the unit of time by two con- 
stant forces would give the relative measures of tlie two forces. 
The absolute measure must be found by comparison with 
some known force, as gravity. Thus let g denote the acceler- 
ation gravity can impart to the unit mass — or any mass, since 
this force increases directly with the mass — in one second^ 
and v the velocity imparted to the same mass in one second 
by the force F, acting with constant intensity ; then, denoting 
the force of gravity by the weight IV oi the mass, we should 
have 

F: W::v:g, 

j? W 

or F = — v. 

g 

So that when IV, g, and v are known, F becomes known in 
pounds pressure. 

[8.] Attraction. — Attraction is a property of bodies 
which causes them to tend toward each other. All matter so 
far as we know is possessed of this property. There are, 
however, several kinds of attraction, as cohesive attraction, 
capillary attraction, magnetic attraction, etc., which are pecu- 
liar to certain bodies under certain conditions, but which 
differ from that which is common to all bodies, in the extent 



DEFINITIONS AND EXPLICATION OF TERMS. II 

and mode of their influence. The force of Attraction exerted 
between masses as they exist in nature, and which is universal 
in its influence, is termed gravitation. The attraction of the 
earth for other bodies is called terrestrial gravity, or simply 
gravity. 

Of the nature of this force we know nothing, except so 
far as it manifests itself to us in its effects. A knowledge 
of the laws of its action, however, furnishes solutions to 
some of the most interesting problems in the system of the 
world. 

The attraction between any two masses, of whatever 
magnitude, is mutual and equal. That is to say, the attrac- 
tion of a mass A for a mass B, is equal to the attraction of B 
for A. 

Conceive the body A to be divided into m elementary 
particles, and B into n such particles, and denote by f the 
attractive force of one particle for another. The attraction of 
one particle of A for the whole of B would then be n X /, and 
the whole attraction of A for B, equal to the attraction of its 
m particles, would be m X n X f 

By parity of reasoning, the attraction of B for A would 
be represented by n X m X /• But «x»X/=».X m X /, 
hence the truth of the proposition as stated. It may be well 
to observe also that the whole force of attraction is equal to 
the product of the elementary particles in the two bodies by 
the attractive force of one particle for another. 

Again, the attraction of any two masses for each other, is to 
the attraction of any other two masses for each other, as the pro- 
ducts of the masses respectively. 

Denote the attraction of A for B by F, and of C for D by 
F v and let m, n, p, q be the number of equal particles consti- 
tuting each mass, in the order given. From the above we 
have 

F : F x :; m X n x / : p X q X / :: *» X n \p X f. 
But 



m A 




n B 


7 = c 


and 


~q~~D 



12 ELEMENTS OE MECHANICS. 

t t mn A X £ -, _ _ ^ 

Ce 77 = 0<^' and F\F x ::AxB:Cx.D, 

as enunciated. 

£tfr. — If £ = D, then F \ F x \\ A : 6". That is, //^ attrac- 
tions of two masses A and C for equal masses, or for the same 
mass, are to each other as A is to C. 

[9.] Elasticity. — By the Elasticity of a body is meant its 
tendency to regain the natural position of its particles when 
they have been disturbed by some extraneous force. When 
the particles of a body are so disturbed in their relative posi- 
tions, as to produce in them a permanent displacement, it is 
said that the limit of elasticity is passed, and the elastic force 
is thereby diminished. 

A body is theoretically perfectly elastic when its regaining 
force, or force of restitution, is exactly equal to the impressed 
force producing the displacement of its particles. 

[10.] Representation of Forces. — It is obvious that 
the intensities of forces may be represented by the abstract 
numbers ; for if the unit force or unit of intensity be repre- 
sented by i or/, the intensities of a double, triple, or quad- 
ruple force, etc., would be properly denoted by 2, 3, 4, or by 

2/, 3A 4A et c 

But both the intensities and directions of forces may be 
represented by straight lines ; for we may take the direction 
of the line for the direction in which the force acts, and if we 
assume a right line of given length to represent the unit of 
intensity, the ratios of forces to each other would be rightly 
given by the ratios of the lines representing them. 

Referring lines representing forces to a scale of equal parts, 
the intensity of any force in relation to others, referred to the 
same scale, will be shown by the number of the equal divisions 
representing the unit force contained in its length. 

[II.] Resultant and Component Forces. — If any num- 
ber of forces be supposed to act upon a mass or material 



DEFINITIONS AND EXPLICATION OF TERMS. 1 3 

particle, free to move in any direction, and these forces are 
not in equilibrium, or in other words do not mutually destroy 
each other, the mass or particle will move in some given 
direction. But this motion might also be due to some single 
force applied to the mass or particle. Such a force, equiva- 
lent in dynamical effect to the combined action of the several 
forces of the system, is called the resultant of the system, and 
the constituent forces the components. 



PART FIRST. 



STATICS. 

[12.] Parallelogram of Forces.— Of the various prin- 
ciples lying at the foundation of the science of Mechanics, 
the principle of the Parallelogram of Forces ma)^ perhaps be 
considered as the most fertile in its results, whilst it affords 
easy solutions to many problems that would otherwise pre- 
sent considerable difficulty. 

This principle may be stated as follows : If a material par- 
tide A (Fig. i) be acted upon simul- 
taneously by the forces A B and A C, 
forming any angle with each other, 
and whose intensities and directions 
are represented by the lines A B and 
A C, then will A D, the diagonal of 
the parallelogram, constructed ipon 
A B and A C as sides, represent in intensity and direction a result- 
ant force, which in its effect would be the exact equivalent of the 
two forces A B and A C. 

[13.] The proofs of the theorem just enunciated are vari- 
ous, and are generally made to depend upon one or more 
preliminary propositions. The following proof is substan- 
tially that given by Boucharlat in his " Elements de Mecan- 
ique," and is made to depend upon three separate proposi- 
tions, which will be given in order. 




Fig. i. 



ST A TICS. 



15 




(A.) If a material particle A (Fig. 2) be solicited by two equal 
forces P and P lt whose intensities and directions 
are represented by the lines A P and A P v then 
will it move in the direction A R of the diagonal 
of the rhombus, of which A P and A P x are sides, 
and A R will be the direction of the resultant of 
P and P v 

This is so obvious as scarcely to need 
any elucidation, yet it may be observed, 
first, that the line of direction of the result- 
ant motion will lie in the plane of the forces 
Pand P v because every reason that could be given for its lying 
on one side of that plane would be equally applicable for the 
other side, and if in either case the reasoning were assumed to 
be conclusive, there would be two motions, which is clearly 
impossible. In like manner, no reason could be assigned for 
the motion of the particle in the direction A m, lying in the 
plane of the forces, that would not also apply to the corre- 
sponding direction A n, on the opposite side of A R. Hence, 
plainly, the direction of the motion would be in the diagonal 
AR, which is therefore the direction of the resultant of the 
forces P and P x . 

(B.) If two forces P and Q (Fig. 3) be applied at the extremi- 
ties of a rigid line A B, then will 
the resultant effect of these forces 
be destroyed by some force R, ap- 
plied at a point C, so situated that 
we shall have 



m 



BC\ AC:: P: Q: 



8 



T 



Fig. 3. 



and hence a force equal to R, applied at C, in the opposite direction, 
will be the resultant of P and Q ; in other words, the point of ap- 
plication of the resultant of the two component forces P and Q will 
be at distances from the points of application of the two forces, 
which are inversely as the forces. 



Q 


oB 






and 


A 


m 


P 


VI 






Q 


710 







ID ELEMENTS OE MECHANICS. 

Assume any point o in the line A B, so situated that we 
shall have 

P Ao 



Make Bn = o B and Am — Ao. 

Hence 

Now if we conceive the line m n, supposed to be homogene- 
ous throughout, to be made up of ponderable parts mo and 
no, and assume the weight of mo to be equal to the force P, 
the line n o will in like manner represent Q. 

But it is obvious that the resultant effect of the weights of 
mo and no, which make up the whole line mn, would be 
equilibrated b)^ a force R equal to the weight of the whole 
line, equal to P-\- Q, applied at its middle point C, in a direc- 
tion opposite to P and Q. 

This point is therefore the point of application of the re- 
sultant of P and Q. But as above, by construction, we have 



p 

~Q 


_Ao 
~ Bo 


_AB — Bo_ \mn- 
A B — A o \mn- 

P BC 
Q~AC 


-Bn 
-Am 


BC 

" AC 



or 



which conforms to the proposition that was to be proved. 

It will be observed that the forces P and Q have been con- 
sidered as tractional forces, or tensions ; but a little consider- 
ation will show that the same method is also applicable to 
them as impulsive forces. For if we conceive each force to 
communicate instantly a certain moving force to the portions 
of the lines mo and no, they would [4, Cor. 1] tend to move 
with the same velocity, and the motion of the whole line 
would plainly be prevented by the application of an impulse 
equal to the resultant of the two given forces at its middle 
point C, in the direction CR. 



ST A TICS. 



17 



(C.) If AB = P and A C = Q (Fig. 4) represent two forces 
applied at the point A of the 
plane m n, then will the dyna- 
mical effect of P and Q be a 
resultant action in the direc- 
tion A O of the diagonal of 
the parallelogram constructed 
upon the sides A B and A C. 
In other zuords, the resultant 
of P and Q will pass through 
O, the extremity of the diag- 
onal A O. 

Produce the line BO 




Fig. 4. 



until Op is equal to AB, and construct the rhombus Oq as 
shown in the diagram. Apply at/ and q, and in the direction 
of the line pq, the two equal and opposite forces Q 1 = Q and 
£> 2 = Q. These forces will mutually destroy each other, and 
hence produce no disturbing effect upon the action of the 
original forces P and Q. 

If now the force P be transferred to the point B in its line 
of direction, making By — AB, its dynamical effect upon the 
plane m n will not be changed, for the same force applied at 
A would still communicate its influence through the rigid 
line of particles lying between A and B to the point B. In 
like manner the force AC — Q may be transferred to the 
point q, taking in direction qx = Q, without change of effect. 
Since Q t and <2 2 have no disturbing effect upon the action of 
P and Q, the effect of the latter will be the same as the effect 
of P, Q v Q and a . But the effect of P applied at B and Q x 
will, from what was shown above, be equal to their resultant 
R, applied at O; for, by construction, 

P: "&■== fi» Op: BO, 

and hence O is the point of application of their resultant, as 
was shown above. 

Again, the resultant of Q applied at q and Q 2 will, as we 
have shown, act in a direction q R x , bisecting the angle Q^qx; 




1 8 ELEMENTS OF MECHANICS. 

and if prolonged in the direction qo, pass through the point 0. 
Therefore, since R and R v the equivalents of the four forces, 
act at the point O to produce a single resultant, the resultant 
effect of P and Q must also pass through O, as was proposed 
to be shown, and hence along the diagonal A O. 

We have thus shown that the resultant of two forces con- 
curring at one point is in the direction of the diagonal of the 
parallelogram, of which the sides represent the magnitude 
and direction of the two forces. 

It remains to be shown that the diagonal, besides giving 
the direction, also represents the mag- ftv* 
nitude of the resultant. 

For this purpose let A P— P and % 
A Q— Q be two forces applied at A 
(Fig. 5). Complete the parallelo- 
gram APRQ. The resultant of P 
and Q will then, from what precedes, take the direction A R. 
Whatever be the value of this resultant, which we may de- 
note by x, it will, if reversed in direction along the line A m, 
hold in equilibrium the forces A P and A Q, or P and Q. 
Again, since this unknown resultant x will then equilibrate 
the forces P and Q, it must, if taken in connection with Q, 
give a resultant AP V equal and opposite to AP or P; AP x is 
therefore the resultant of the unknown force x and Q. 

Joining P 1 and Q by a straight line, and drawing P 1 R 1 
parallel to A Q, since P x Q is parallel to A R or A R„ the 
figure A QP 1 R 1 will be a parallelogram. Hence, since a force 
represented by A R, in the direction A m is the only force 
which, taken in connection with the force Q, will give a result- 
ant lying in the known direction A P 1 of the resultant of Q and 
x, it follows that A R x is equal to x, the resultant of P and Q. 

Again, since both AQP l R 1 and AP.QR are parallelo- 
grams, we have 

x = AR 1 =QP 1 =AR. 

Hence A R, the diagonal of the parallelogram constructed 
upon the lines A P and A Q representing the component 



ST A TICS. 19 

forces, will represent both the direction and magnitude of 
the resultant ; which conforms to the statement of the theorem. 
A little consideration will show that there is nothing in the 
above reasoning that will not apply to impulsive or constant 
tractional forces. 

[14.] If we conceive a particle to be impelled by two 
forces, making any angle with each other, one of which is 
a single impulse and the other a succession of impulses, at 
equal intervals of time and parallel to each other, its motion, 
on account of the repeated deflections caused by the succes- 
sion of impulses, would be a broken line. 

Should the intervals of time become infinitesimal, thereby 
rendering the second a constant force, the deflections would 
take place at every instant of time, and the broken line would 
become a curve. 

Since any two constantly acting forces, constant or vari- 
able, may be assumed to be made up of an indefinite number 
of equal or unequal impulses, acting at intervals of time that 
are infinitesimal, the path pursued by a particle, acted upon 
by two such forces following the diagonals of the indefinitely 
small parallelograms whose sides are represented by the im- 
pulses corresponding in time, would plainly be a curve either 
regular or irregular in its form according to the law of the 
forces or the values of the various impulses ; observing, how- 
ever, that the curve becomes a straight line when both forces 
become constant. 

[15.] Triangle of Forces. — If three forces, whose direc- 
tions and intensities are represented by the sides of a triangle 
taken in order, act upon a material particle, they will be in equi- 
librium, and no motion will ensue. 

Let AD, DC, and CA represent the " ^ 

directions and intensities of the given 
forces, and complete the parallelogram 

ABCD. Supposing the force D C to DX ~ " ""^ 

preserve its direction, with its point 01 

application at A, it would evidently be equivalent to a force 




20 ELEMENTS OF MECHANICS. 

represented by A B. But A D and A B give a resultant A C. 
Hence, since the given force C A is equal to and directly 
opposed to this resultant, it will destroy its effect, and the 
given forces will be in equilibrium. A little consideration 
will show that the converse of this proposition is also true. 

[16.] Polygon of Forces.— If any number of forces, rep- 
resented in direction and intensity by the sides of a polygon taken 
in regidar order, act upon a material particle, the forces will be 
in equilibrium, and no motion will ensue. 

Let AB, BC, CD, DE, and EA represent the directions 
and intensities of the given forces, and A the point at which 
each force preserving its proper direction is supposed to be 
applied. 

Draw the diagonals A C and A D. The forces A B and 
BC, applied at A, give a resultant A C. 
But A C and CD, applied at A, give the 
resultant AD; and finally^/? and DE 
give A E as their resultant : which is there- 
fore the resultant of all the given forces 
except EA ; and since E A is equal and 
directly opposed to A E, it will destroy its 
effect, and produce equilibrium. 

It is obvious that the same mode of proof may be adopted 
whatever be the numbers of the given forces, and it may be 
observed, also, that there is no necessity that all the forces 
should lie in the same plane, for it is evident that in whatever 
plane A B and B C lie, for example, A C will still be their re- 
sultant, and in like manner A D and A E will be the other 
resultants of the system. 

It may also be shown that if any number of forces con- 
curring at one point are in equilibrium, and a figure be 
formed by starting from the point and drawing, in successive 
order, lines to represent in magnitude and direction the forces, 
also taken in order, the figure thus formed will be a closed 
polygon. 




STATICS. 21 

For if this be not true, the line representing the last 
force taken will not close the polygon, but some other line 
will, and this latter, by the proposition just proved, will rep- 
resent in magnitude and direction a force which will hold the 
remaining forces of the system, when applied at the given 
point, in equilibrium. But by hypothesis the last force taken 
of the given system will, in connection with the remaining 
forces of the system, produce equilibrium ; hence the force 
represented by the closing line of the polygon can be no other 
than this force, and the truth of the proposition becomes ap- 
parent. 

[17.] The principle of the Polygon of Forces furnishes at 
once a method given by Leibnitz for determining by con- 
struction the resultant of any number of given forces concur- 
ring at a point. For from any assumed point draw a straight 
line to represent the magnitude and direction of the first of 
the given forces ; from its extremity a line to represent the 
second force ; from the extremity of the latter a line to repre- 
sent the third given force ; and so on until all of the forces 
given have been employed. 

If, then, the figure so formed be a closed polygon, the 
forces will be in equilibrium, and the resultant equal to zero ; 
but if not, the line joining the extremity of the line represent- 
ing the last force and the assumed point will represent a force 
that would equilibrate the given forces, and hence, if reversed 
in its direction, would be the required resultant of the given 
system. 

EXAMPLE. 

Four forces are given, in magnitude and direction as fol- 
lows: 

N. 20° W., io lbs. traction. 
N. 30 E., 15 " 
S. 50 E., 12 " 
South, 14 " 

It is required to find, by construction, the magnitude and 
direction of their resultant. 



22 ELEMENTS OF MECHANICS. 

The student may also observe that since the forces em- 
ployed are directly proportional to the lengths of the lines 
representing them, the case is precisely analogous to that of 
supplying the last line in a survey when the bearings and 
distances of the other sides are given ; which can also be 
effected by a graphical construction. Let the student, how- 
ever, solve the example by finding the difference of Latitude 
and Departure of the given lines representing the forces, and 
calculating the bearing and length of the closing line of the 
polygon, without instruments. The direction of the result- 
ant will obviously be the reverse bearing of this line, whilst 
the length will represent the magnitude of the resultant. 



Applications of the Foregoing Principles. 

GRAPHICAL SOLUTION OF PROBLEMS. 

[18.] Example i. — A point A is solicited by three forces, 
A m, A n, and A o (Fig. 8), whose intensities are represented by 
the length of the lines, as in the preceding articles. It is re- 
quired to find a single resultant force, the equivalent of the 
three given forces. 

Constructing the parallelogram AmCn with the sides A m 
and A n, we shall have A C, a force 
represented by the diagonal, as the 
equivalent of the two combined 
forces A m and A n. In like manner 
the forces A C and A o, when acting 
together, would produce a resultant 
A B, represented by the diagonal of 
a parallelogram, constructed with 
these forces as sides. The force A B 
is therefore the resultant of A C and FlG - 8 * 

A o, or, which is the same, of A m, A n, and A o. The same 
method may be extended to any number of forces. 

The magnitude of the forces being laid off from some 




STA TICS. 



23 




Fig. 



assumed scale of equal parts, the magnitude of the resultant 
will be found by referring it to the same scale of parts as in 
all other cases of the graphical solution of problems. 

[19.] Ex. 2. — To find that element of the force of gravity 
which constitutes the vibrating force of a simple pendulum at 
any point P (Fig. 9) in its arc of vibration. 

Laying off P W to represent the weight of the pendulum, 
acting in a vertical direction, and resolving it a 
into the two forces Pm and Pit, the one per- 
pendicular to A P, and the other in its line of 
direction, we may suppose the pendulum P 
to be influenced simply by these two forces, 
of which the force Pm alone tends to produce 
vibration, whilst Pn produces simply a tension 
of the suspending cord A P. To estimate the 
value of the force Pm, it must of course be 
referred to the same scale of measurement 
adopted in laying off P W, to represent the weight of the pen- 
dulum. 

Problem. — The suspending cord of a simple pendulum 
makes an angle of 50 with a vertical line through the point 
of suspension. Required the force tending to produce vibra- 
tion, when the weight of the pendulum is 10 ounces. 

[20.] Ex. 3. — A material particle is solicited by three forces 
represented by the three edges of a rectangular parallelopipe- 
don. Construct the diagram, and show that the resultant 
will be the diagonal of the parallelopipedon. 

[21.] Ex. 4. — A body (Fig. 10) is placed upon an inclined 
plane A C. Let it be required 
to determine the force urging 
the body down the plane, and 
also the perpendicular pressure 
upon the plane. 

Laying off from a scale W, 
to represent the given weight 




24 ELEMENTS OF MECHANICS. 

of the body, it may be resolved into the two forces of and op, 
respectively parallel and perpendicular to the plane A C. 
The force op, producing a simple pressure of the body upon 
the plane, does not tend to produce motion, and the force of 
alone represents the force down the plane. 

Denoting this force by /, the height of the plane by h, and 
its length by /, we have, from similar triangles o Wf and ABC, 
the proportion 

/: Ww h\ /, 

whence /.== —j-', 

or, calling radius unity, we have 

fz= W. sin C. 

Should it be required to apply a force to prevent the 
motion of the body down the plane, under the supposition 
that there is no friction, it would obviously be necessary to 
apply one equal to and directly opposed to the force /. 

It will be seen from the above equation that the forces 
required to move the same body up different inclined planes, 
when applied in directions parallel to the lengths of the planes, 
would be proportional to the sines of the angles of elevation of 
the planes. 

Since the value of the component of would not change 
with the position of the body upon the plane, it acts as a con- 
stant force in moving the body, which is to the force of gravity 
or weight of the body as the height of the plane is to its 
length. 

Referring again to the diagram, and supposing the weight 
of the body o l to be resolved into the two components op^ and 
of, respectively perpendicular to the plane and parallel to its 
base, there would result, from similar triangles, 

W x \f\\b\ h, 

W x h 



ST A TICS. 



25 



Since, also, 
we have 



7- = tan c, 
f x = W x tan c, 



and the force required to prevent sliding down the plane, 
when there is no friction and the force is applied parallel to 
the base of the plane, is proportional to the tangent of inclina- 
tion. 

In fact, however, the increase of pressure upon the plane 
with the decrease of its angle of elevation would increase the 
friction, so that, in practice, the actual force necessary to pre- 
vent motion of the body down the plane would decrease more 
rapidly than indicated by the above formulae, the friction 
alone becoming sufficient to prevent motion as the angle of 
elevation decreases. 



An inflexible rod 
A * 




Fig. 11. 



[22.] Examples for Solution. — (1.) 
BF (Fig. 11) resting upon a fulcrum F 
in the face of a vertical wall, with a given 
weight ^suspended at its extremity B, 
is kept in position by the horizontal line 
A B. Supposing the angle A F B to be 
equal to 50 , and the weight to be equal 
to 20 lbs., required first the tension of the 
cord A B, and secondly the pressure 
upon the rod, the latter being supposed to be possessed of no 
weight. 

Explain also the mode of solving the example when the 
rod is supposed to be of given weight. 

(2.) A beam ^ B (Fig. 12), suspended 
by two parallel cords A D and B C, is 
prevented from hanging with the cords 
vertical, by a force FG, applied at F. 

Given the weight of the beam equal 
to 20 lbs., the angle BCD — 50 , and 
A F G = 20 , it is required to find the 
value of the force F G. 




Fig. 12. 



26 ELEMENTS OF MECHANICS 

(3.) A strong cable or iron rod is supposed to span a stream 
which has considerable current, and to be firmly attached to 
two trees or fixed abutments on the opposite banks. From 
the two ends of a ferry-boat in the stream are ropes that 
attach to a pulley that rolls upon the cable. 

Show, by the composition and resolution of forces, that if 
the boat is made to assume a position inclined to the current 
by means of a windlass, to wind up or let off the rope at one 
end, it can be made to cross and recross the stream by the 
action of the current alone. 

It may first be shown that the tension upon the ropes, the 
weight of the boat, and the buoyant force of the water will 
give a resultant acting directly up-stream. The pressure of 
the current may be resolved into one perpendicular to the 
gunwale, and one parallel to it, which latter has no appreciable 
effect. The perpendicular component may then be combined 
with the resultant referred to above, giving a force transverse 
to the stream. 

(4.) If O be a point within a triangle ABC, and lines be 
drawn to the vertices A,B, C, and also to the middle points of 
each side, then will the system of forces represented by the 
first three lines, acting at the point O, be equivalent to that 
represented by the last three. 

It should be borne in mind that the diagonals of a paral- 
lelogram mutually bisect each other. 

[23.] If two forces whose lines of direction meet in a point 
are held in equilibrium by a third force, this latter force must 
also pass through the same point ; for it is the same as the 
resultant of the two given forces reversed in direction ; but 
this resultant passes through the given point, and therefore 
the equilibrating force also. 

PROBLEM. 

Two homogeneous spheres, whose radii are respectively 
2 and 3 inches (Fig. 13), and the weight of the smaller 8 pounds, 
are suspended at the point A by two cords, A B = 6 and 



STA TICS. 



27 




Fig. 13 



A C — 8 inches. It is required to show how, in case of equi- 
librium, the tension of the two cords may be found. 

First, if we conceive the spheres to be rigidly connected 
by the line B C joining their centres, 
it is obvious that the position of rest, 
or equilibrium, will not be disturbed. 

This equilibrium results from the 
action of three forces : the tensions 
upon the two cords, and the weights 
of the two spheres acting through f* 
their centre of gravity, supposed to be / 
at g. As the weight acts in a vertical \ 
direction, and in case of equilibrium, \ 
from what precedes, must pass through 
the point of support A, the centre of 
gravity g must lie in the vertical line Ag, passing through A. 

It is obvious, also, [13] that the pointy will be at distances 
from B and C inversely as the weights, and hence inversely 
as the volumes of the spheres ; or, since these volumes are as 
the cubes of the radii, as 8 to 27. Hence if the line B C = 5 in. 
be divided into two parts, to each other as 8 to 27, the point 
g is determined, and Bg and Cg become known. 

Now let A g = W, be the combined weight of the spheres, 
easily found, and draw gm parallel to CA t and mA along 
BA. 

By the principle of the triangle of forces [15], the lines Ag r 
gm, and mA lying in the direction of the three forces produc- 
ing equilibrium at A are proportional to those forces. Hence, 
since A g is assumed to represent one of these forces, or the 
weight W, we have gm = t and mA — t x for the tensions upon 
the cords A C and A B. Designating the angles as shown in 
the diagram, we have, from principles of trigonometry, 



t : t x : W :: gm : mA : Ag :: sin /? : sin a : sin A mg 
= sin Bmg, 
or / : i y : W : : sin J3 : sin a : sin (a + /?). 



28 



ELEMENTS OF MECHANICS. 



Now in the triangle ABC, all the sides being given, the 
angles B and C may be found. In the triangles A Bg and 
A Cg we then have, in each case, two sides and the included 
angles to find the angles a and /?. 

Finally, in the triangle A gm we have Ag = W, and a and 
ft known, to find t and t l9 the required tensions of the cord 
in case of equilibrium. 



Composition and Resolution of Forces.— Concurring 

Forces. 

DETERMINATION OF THE RESULTANT OF ANY NUMBER OF 
CONCURRING FORCES IN ONE PLANE. 

[24.] Having shown [13] that a force represented in direc- 
tion and intensity by a straight line may be resolved into two 
other equivalent components, represented by the sides of a 
parallelogram, of which the line representing the given force 
is the diagonal, we are furnished with a convenient method of 
determining the resultant of any number of concurring forces, 
acting in the same plane, and whose intensities and directions 
are given by referring them to a set of rectangular axes. 

Let P, P v P„ P s} etc. (Fig. 14), be the given forces concur- 
ring in the point A> and represent 
their intensities and directions re- 
spectively by AP, AP iy AP„ etc. 
Let also the angles which the direc- 
tions of these forces make with the 
axis of X or A X, reckoned in the 
direction X Y, through 360 , be de- 
noted respectively by a y a v a„ etc., 
and those formed in like manner 
with the axis of Y or A K, reckoned 
in the same direction, by /3, j3 v fi v etc. 

Resolving A P into the two equivalent components Ax 
and Ay, lying in the axes of X and F, we have, in accordance 
with our notation from trigonometrical principles, 




Fig. 14. 



3TA TICS. 29 

A x — A P cos a = P cos a 

and Ay = A P cos fi — P cos 

for the values of the components in the direction of the co- 
ordinate axes, due to the force P. 

In like manner, for the components due to P lt P„ P v etc., 
we obtain the similar expressions 

P 1 cos oc v P l cos fi 1 ; P 2 cos a„, P 2 cos fi 2 ; etc. 

Hence, whatever be the number of forces applied at the 
points, each one of them may obviously be resolved, accord- 
ing to the same method, into equivalent components acting in 
the direction of the axis. 

If we assume the components acting in the direction A X 
and A Fto be positive, those acting in the opposite directions 
A X x and A Y v since they oppose the former, should be con- 
sidered as negative. In conformity with this hypothesis it 
will be seen by a slight inspection of the diagram that if the 
original forces P, P v P 2 , etc., be assumed to be positive, the 
signs of the cosines in the different quadrants will be such as to 
give to the components in the direction of the axes their 
proper signs, as indicated above. It will be easy for the 
student to confirm this statement by reference to the diagram, 
observing the direction of the various components, and the 
signs given to them by their trigonometrical expressions given 
in the foregoing equations. 

Denoting by X the algebraic sum of all the components 
lying in the axis of X, and by Y the sum of those lying in the 
axis of Y, we shall have the equations 

X = P cos a 4- P l cos a l + P. z cos a„ -|- etc. 
and Y = P cos + P, cos ft, + P, cos ft + etc., 

in which equations, as stated above, the cosines are supposed 
to be affected with their proper signs. 

Thus the two forces X and F, equal respectively to the 
algebraic sum of the components in the two axes, are the 



30 ELEMENTS OF MECHANICS. 

resultants or equivalents of the whole system of forces, and 
may be substituted for them without change of mechanical 
effect. But the resultant of these two forces, acting in the 
axes of X and F at right angles to each other, is the diagonal 
of a rectangle of which they form the sides. Hence, calling 
this resultant R, we shall obviously have 

jp = X" + F 2 , and R = VX 2 + F 2 
for the magnitude of the resultant. 

To determine its direction in the plane of the original forces, 
or of the co-ordinate axes, let a and b denote respectively the 
angles it makes with the axes of X and F, reckoned as before. 
We shall then have, from trigonometrical relations. 

X A 1 Y 

cos a = r=r and cos b = -=-, 
A A 

which equations, since X } F, and R, are already known from 
the preceding equations, will determine the value of the angle 
the resultant makes with the one or the other of the co-ordi- 
nate axes, which is sufficient to fix its position. 

It will of course be understood that the magnitudes of the 
given forces, as well as their directions, or the angles they 
form with the assumed axes, are supposed to be known. 

If the system of forces be in equilibrium, its resultant must 
evidently be equal to zero. Hence, also, 

A 2 = X" + F 2 = o. 

But since every square is essentially positive, this equation can 
only be true when both X—o and F= o, showing that the 
algebraic sum of all the components in either of the co-ordinate 
axes must be equal to zero. 

This would plainly appear, also, from other considerations ; 
for if we conceive that the components lying in the axes do 
not in each case reduce to zero, the two remaining forces, the 



ST A TICS. 3 I 

resultants of the components in the two axes, would act at right 
agles to each other, and could not therefore produce equi- 
librium. 

Representing by the Greek symbol 2 the algebraic sum of 
the whole system of quantities of the form of that to which it is 
prefixed, the preceding equations for the algebraic sum of the 
components in the directions of the respective axes may be 
written 

X=2Pcosa and Y= 2Pcos/3. 

Hence, also, 

R* = JP + F 2 = {2Pcos af + {2 Pcos /3)\ 

Note. — Since, both in magnitude and sign, cos fi = sin a, 
cos /3 1 = sin a v cos /? 2 = sin a v etc., as will become manifest 
by a careful inspection of the diagram, we may write 

X — 2 P cos a, Y=2Psina, 

and R 2 = (2 P cos of + (2 P sin a)\ 

It should be observed that in the above discussion trigo- 
nometrical formulae have been employed, in which radius is 
supposed to be equal to unity ; and hence, in the application 
of the derived equations to the solution of particular ex- 
amples, the cosines or sines of the various angles involved in 
them may be taken from the tables of natural sines, cosines, 
etc. 

The following examples will serve to illustrate the formulae 
deduced. 

[25.] Examples. — (i.) Four forces, P, P v P„ P z , whose mag- 
nitudes are represented by the numbers 2, 3, 4, and 5 respec- 
tively, make with the axis of X the angles a=$°, 0^ = 30°, 
a 2 = 50 , and a 3 = 140 , and with the axis of Y the angles 
f3 = 85 , fi l = 6o°, /? 2 = 40 , and /?, = 50 . It is required to 
determine the magnitude and direction of the resultant when 
the forces concur at the same point. 



32 ELEMENTS OE MECHANICS. 

Referring to a table of natural sines and cosines, and 
taking out the values of the cosines of the given angles, 
we have 

X—P cos a -f- P l cos or, + P 2 cos ^ 2 -\- P 3 cos <r 3 = 2 X .99619 
+ 3X .86603 + 4 X .64279 - 5 X ./6604 = 3.53 143 ; 

Y=Pcos fi + P, cos A +P, cos /?„ + P, cos /i 3 = 2 x .08716 
+ 3 X .50000 + 4 X. 76604 + 5 X .64279 = 7 .95243. 

Substituting these values of X and Y, we have for the 
magnitude of the resultant, 



R = I'J 2 + F s = 8.62204 ; 
and for the angles it makes with the co-ordinate axes, 

cos a = ~= M3I43. = <38638 
.# 8.62204 ° ° 

» 

j , Y 7.95243 

and cos £ = — = J ^/ ^° = . 92234. 

R 8.62204 y ^ 

Consulting the tables for the values of the angles whose 
cosines approximate most nearly to these values, we find 
a = 6y° 16' and £ = 337 i&, the direction of the resultant 
lying in the first quadrant, since both cosines are positive. 

Thus the direction of the resultant becomes known. The 
relation existing between the values of a and b, ascertained 
by considering the diagram when the quadrant in which the 
resultant lies is determined, furnishes also a test of the accu- 
racy of the results. In the present instance, for example, the 
resultant lying in the first quadrant, the difference between 
360 and the angle b must be the complement of the angle a\ 
as will be found to be the case. If it fell within the second 
quadrant, the difference between 180 and the angle a would 
be the complement of b. The student may consider the other 
cases for himself. 



ST A TICS. 



33 



(2.) The four forces are P — 5, P x — 4, P 2 = 6, and P 3 = 6, 
and the angles made with the axis of X are respectively 
a = 24 , a x = 50 , or a = 1 35 , and or, — 310 . To find R, a y and b. 

(3.) Three forces, represented by the numbers 7, 10, and 
12, meet at a common point, and make with each other angles 
of 120 . Required the magnitude of the resultant, and the 
angle it makes with the least force. 

For convenience take the least force to coincide with the 
axis of X. 

Note. — It is obvious that a graphical solution of the above 
examples would present little difficulty, and if the radius of 
the ordinary logarithmic tables were introduced into the for- 
mulae for cos a and cos 0, the same results could be obtained 
by logarithmic methods. 



Composition of Two Concurring Forces in One Plane. 

SECOND METHOD. 

[26.] Let P and P x (Fig. 15) be two forces concurring at 
the point A f and making with the axis of 1 
X angles respectively equal to Fand V v Y 
and denote by a the angle the forces 
make with each other. We shall then 
have, from principles of Geometry, 




Fig. 15. 



J? = A P + A P^ -f 2 A P, X P y n y 

or R* = P 2 + P: + 2P 1 xP 1 n. 

But P x n — P cos a, 

and 
R* = P* + P* 4- 2 PP l cos a = P* + P? + 2 PP X cos ( V - V x \ 

If the force P x be supposed to coincide with the axis of X, 
then 

V x = o and V— a, 

whence R — VP 2 + P* + 2PP X cos a; 



34 ELEMENTS OF MECHANICS. 

that is to say, the resultant of two forces acting at the same point 
is equal to the square root of the sum of the squares of the two 
forces, plus twice their product, into the cosine of their included 
angle. 



If V=a = o, we have R = VP'+P^+zPP^P+P,. 
If V=a = 9 o°, we have R*=P* + P*; 
and if V= 180 , 



cosF= — i, and R= VP* + P* - 2PP X = P- P v 

These results are such as might have been anticipated, for, 
under the first supposition, both forces act in the same direc- 
tion, and their resultant should be equal to their sum ; in the 
second case, the resultant R becomes the diagonal of a rect- 
angle, of which P and P 1 are the sides, whence R* = P 2 -\-P? ; 
and under the last hypothesis, Pand P x act in opposite direc- 
tions in the same straight line, and hence, as should be the 
case, the resultant is equal to their difference. 

To determine the direction of the resultant R we may 
have recourse simply to the equation of the preceding article, 

X PcosV+P.cosV, 

whence cos a — -= - = 5 . 

K A 

In the following examples it will be convenient also to use 
the table of natural sines and cosines. 

(1.) Two forces applied at the same point, and equal 
respectively to 10 and 16, make an angle with each other 
equal to 30 . It is required to find the magnitude of their 
resultant, and the angle it forms with the lesser force, assumed 
to coincide with the axis of X. 

(2.) Two forces, assumed equal to 8 and 12 respectively, 
produce a resultant equal to 16. It is required to find the 
angle they make with each other, and also the angle between 
the resultant and the greater force. 




STATICS. 35 

[27.] If three forces, P, P lf and R (Fig, 16), be supposed 
to hold the point A in equilibrium, 
it is evident that the force R must 
be equal and opposite to R v the 
diagonal of the parallelogram A B, 
and the resultant of P and P v 

But from the known relation 
which exists between the sides of 
a plane triangle and the sines of the fig. 16. 

angles opposite, we have 

AP: PB : AB :: sinABP: s'mBAP: sinAPB, 
or P: P x \ R\\ sin BAP, : sin ^ P : sin CA P x . 

But since the sines of angles are equal to the sines of their 
supplements, 

P: P x \ R :: sinP.AR : sinPAR : sinP^P; 

whence we conclude that when three forces acting at a point are 
in equilibrium, each force is proportional to the sine of the angle 
included between the directions of the other two. 

The converse of this proposition, however, without a fur- 
ther imposed condition that each force shall lie without the 
angle formed by the other two, is not true ; for it is obvious 
that if the force R, for example, be reversed in its direction, 
becoming R v and the supplements of the angles P^AR and 
PA R be substituted in the proportion last obtained, each of 
the forces P, P v and R x would be proportional to the sine of 
the angle included between the other two, and yet not be in 
equilibrium. The same could be shown if either force were 
reversed in its direction, and under this supposition the resul- 
tant of the system would obviously be double the value of the 
force so reversed, and coincident with this reversed direction. 

EXAMPLES. 

(i.) A material point is held in equilibrium by three given 
forces, P, P x , and R. It is required to find the angles the 



36 ELEMENTS OF MECHANICS. 

forces make with each other. Denoting the angles between 
P and P v P and R, and P x and R by a, ft, and y respectively y 
we have from the preceding article, since R must equal the 
resultant of P and P x in magnitude, the equation 

R> = P 2 + P 1 * + 2 PP 1 cos a, 



and cos a 



2ffi 



from which the angle a becomes known. 
From analogy we can also write 

^ /> = ^^ and ^ Y = Yp^ R • 

Or, having found the value of a, and thence of sin a, we 
have, from proportions already deduced in the present article, 

P 1 : R : : sin fi : sin a and P : R : : sin y : sin a, 

P P 

or sin /3 = — - .yzVz a and jzt* ;/ = — sin a, 

R R 

the second members of these equations being known. 

(2.) Let P— 10, P 1 — 12, and R = 16, to determine the 
angles made by the forces with each other. 

(3.) Two given forces, equal to 9 and 15, make an angle 
with each other equal to 64 . Required the third force 
necessary to produce equilibrium, and the angles it makes 
with the given forces. 



Determination of the Resultant of any Number of Con- 
curring Forces not in the Same Plane. 

[28.] By reference to [20, Ex. 3], which the student is 
supposed to have solved, it is found that any force, as A P 
(Fig. 17), may be resolved into three others, A b, Ac, and A d y 
represented by the three edges of a parallelopipedon, of which 



ST A TICS. 



37 




the given force A P is the diagonal. Assuming the three rect- 
angular axes XX X , ZZ X , and Y Y x to coincide with the three 
edges of the parallelopipedon, and denoting by a, /3, and y 
respectively the angles 
formed by the force P with 
the axes X, Y, and Z, we 
shall have, from the right- 
angled triangles APb,APc, 
and A Pd, the equations 

A b = Pcos a, 
A c = Pcosy, 
Ad=Pcosfi. 

But A b, Ac, and A d are 
the components in the di- 
rection of the axes, equiva- 
lent in effect to the single 

force P, and may therefore FlG I7 

be substituted in its stead. 

It is obvious that whatever be the number of forces ap- 
plied at A, each one of them may in like manner be resolved 
into three component forces acting along the axes without 
change of effect. 

If we regard all the components in the axes in the direc- 
tions AX, AZ, and A Y as positive, and those acting in the 
opposite directions as negative, and consider the angles formed 
by each force with each one of the axes to be measured from 
AX, AY, and A Z through 180 in the plane of the axis and 
the line representing the force, then, as we shall see, the com- 
ponents in the direction of the axes will, in virtue of the signs 
of the cosines of these angles, assume the signs properly be- 
longing to them, in whatever direction the force may act. 
This will become apparent only by considering the diagram, 
and conceiving lines representing the directions of the forces 
to be drawn in each of the eight triedral angles formed by the 
intersection of the three co-ordinate planes passing through 
the axis. 



38 ELEMENTS OF MECHANICS. 

If, for example, the force lies in one of the four triedral 
angles to the right of the plane X Y Z x , its component along 
the axis of X will be positive, but if on the left of the same 
plane it will be negative; and this would plainly be indi- 
cated by the signs of the cosines of the angles formed with 
the axis of X; for under the first supposition all these angles 
would be acute and their cosines positive, and under the sec- 
ond they would be obtuse and their cosines negative. 

Similar observations might be made in reference to the 
planes X ZX X and X YX X , since it is easily seen that all forces 
lying in the triedral angles in front of the plane XZX X would, 
on account of the signs of the cosines of the angles made 
with the axis A Y, give positive components along the axis of Y y 
whilst those lying behind the plane XZX X would give negative 
values for the components, the cosines of the angles made with 
the axis Y being negative. 

So, also, the components of all forces above the plane 
X YX X in the axis of Z would be rendered positive, and those 
of forces lying below the same plane would be negative. 

[29.] If, therefore, there be any number of forces P,P X ,P„ 
etc., concurring in the point A, and having all possible direc- 
tions, there will result from what precedes, for the algebraic sum 
of the components in the direction of the three axes, which 
we may denote by X, Y, and Z, the following equations, viz. : 

X = P cos a -f- P x cos a x -\- P 2 cos a 2 -f- etc., 
Y=Pcos/3 + P l cos ft \ -f P 2 cos ft 2 -f etc., 
Z = P cos y -\- P x cos y x -[- P % cos y^ -f- etc. ; 

or, adopting the same notation as in [24], 

X=2Pcosa, Y = 2 P cos ft, Z=2 Pcos y. 

[30.] The directions and intensities of the forces being 
given, the values of X, Y, and Z, the resultants of the com- 
ponents lying in each axis and representing the three edges 



ST A TICS. 39 

of a rectangular parallelopipedon, of which the diagonal is the 
resultant of the whole system, would become known, and we 
should have 



R* = X 2 + Y 2 + Z\ or R = YX 2 + Y 2 -\- Z\ 

If we denote by a, b, and c the angles the resultant makes 
with the three axes A X, A Y, and A Z, we have 

X A Y A Z 

cos a = -=? , cos b — 77, and cos c — -77, 
K K K 

which equations will make known the direction of the result- 
ant in reference to the fixed axes. 

By squaring and adding the last three equations, we have 

cos a -f- cos o-f-cos c = — = 1, 

a well-known formula. 

[31.] If the system of forces be in equilibrium, the result- 
ant R must be equal to zero. Whence 



YX 2 + Y 2 + Z 2 = o, or X 2 + Y 2 + Z 2 = o, 

which, as every square is essentially positive, can only be true 
when X, Y, and Z, are separately equal to zero. 

That is to say, the algebraic sum of the components in the 
direction of each axis must be equal to zero : which is obviously 
true, without reference to what precedes ; for if there could be 
any remaining forces in the three axes due to the fact that the 
components in the respective axes did not destroy each other, 
they would not all three act in the same plane, and could not 
therefore produce equilibrium ; nor if the components in any 
one axis destroyed each other, could the resultant forces in 
the remaining two, acting at right angles to each other, pro- 
duce equilibrium. 

[32.] If the angles that any given force makes with two of 
the axes be known, it will be easy to determine the value of 



40 ELEMENTS OF MECHANICS. 

the angle made with the remaining axis ; for, from [30], if we 
suppose a and b to be the given angles made by any force 
whatever with the axes of X and Y> and c the angle made 
with the axis of Z, we shall have 



cos c 



= ± Vi — <;<?/# — £w 2 ^, 



from which the value of c becomes known, and the direction 
of the force determinate with regard to one of two positions, 
according as the plus or minus value of cos c is taken. The 
plus value applying to a force above the plane of X Y, and the 
minus sign to one below the same plane ; the angles they 
make with the axis of Z being supplements of each other. 

EXAMPLES. 

(1.) Required the magnitude and direction of the resultant 
of three forces acting at right angles to each other, whose 
intensities are denoted by 7, 8, and 10 lbs. respectively. Let 
the forces be supposed to coincide in direction with the axes. 

(2.) The resultant of three forces at right angles to each 
other is 100 lbs., and the forces are to each other as the num- 
bers 7, 8, and 10. Required their intensities, and the angles 
the resultant makes with the axes of X and Y, when the first 
force makes angles of 30 and 6o° with these axes, and the 
second force angles of 6o° and 1 50 with the same axes respec- 
tively. 

(3.) Three forces P, P v and P w acting at the origin of co- 
ordinates, make angles with the axes of X and Y as follows : 

ForP a = 48°, p— 50 , and cos y positive. 

" P x ^=60°, A= 130 , " 

" P* " 2 =30°, A= 85 , " " negative. 

Supposing the resultant to be equal to 100, and to lie in 
the axis of X, let the student explain in what manner the 
values of the three forces can be determined, by deriving the 
necessary equations, and solving them if deemed expedient. 



ST A TICS. 



41 




Non-Concurring Forces. 

PARALLEL FORCES IN THE SAME PLANE. 

[33.] Let A M = P and BN = P x (Fig. 18) be two parallel 
forces applied at the points A and B 
of the plane X Y. Apply also at A and 
B the two equal and opposite forces A O 
and B O x , which, since they mutually de- 
stroy each other, do not change the effect 
of the original forces P and P x . Com- 
pleting the parallelograms O M and O x JV, 
the resultants A R and B R 1 are therefore 
the equivalents of the forces P and P x . 

Prolonging RA and R X B to meet at some point V, take off 
VD = BR, and VF=AR. Draw VG parallel to the direc- 
tion of the forces P and P v and complete the parallelograms 
KE and HC, similar and equal to the parallelograms O x A r and 
OM. If we now suppose the forces AR and BR X to change 
their points of application to V, in their lines of direction, the 
mechanical effect upon the plane X Y will obviously be the 
same, for in any case we can conceive a rigid line of particles 
between^ and V, and B and V, to communicate the two forces 
to that point. Hence, following up the process, we have P 
and P x equivalent to A R and B R v equivalent to VF and 
VD, and, since VH and VK destroy each other, equivalent 
to VE and VC; that is, equivalent to P and P x acting in the 
same direction VG, or to (P-\-P x ) transferred to the point G, 
and acting parallel to the original forces. Hence the resul- 
tant of the two forces is equal to their sum, or R = P-\-P x , and 
acts in a parallel direction. 

Moreover, since by construction VG is parallel to P and 
P x , similar triangles give 



VC : CD :: VG : GB 
whence, since EF — CD, 



and 



VE:EF;i VG: GA; 



VE: VC :: GB: GA, 



or 



P:P :: GB : GA. 



4 2 ELEMENTS OF MECHANICS. 

Hence we conclude, first, that the resultant acts in a direc- 
tion parallel to the given forces, and is equal to their sum; and, 
secondly, its point of application is at distances from A and B in- 
versely proportional to the forces applied at those points. 

Or, to render the proposition still more general, the last 
proportion becomes, by composition, 

(P+P x ) : P:: (GB+GA) : GB 

and (P+P 1 ) '•P 1 '-: (GB + GA): GA; 

or, which is the same, 

R\P\\AB\GB and R\P X \\AB\GA\ 

also we have above, 

P: P X \\GB\ GA. 

Hence, from these three proportions we conclude, gener- 
ally, that any two of the three forces P, P v and R, their resultant, 
are to each other inversely as their distances from the third. 

If we take any two of the last three proportions in connec- 
tion with the equation P-\-P 1 = R, they will contain six un- 
known quantities, the three forces and the three distances ; 
and when any three of these quantities, except the three dis- 
tances, are given, the remaining three may be easily found by 
the ordinary methods of elimination. 

That the three distances being given will not determine 
the forces may be easily inferred from the fact that the point 
G would remain the same for any two forces bearing the same 
ratio to each other as P and P v 

The student will observe that the main proposition of this 
section had already been established by an independent method 
in section [13, B], 

EXAMPLES. 

(1.) Let P— 10, AB = 12, and A G — 5, to find the result- 
ant R, the force P lt and the distance B G. 

(2.) Let R = $6, P=2o, and A G = $. Required A B y 
B G, and P v 



STA TICS. 43 

[34.] Since the resultant R of the two forces P and P x 
would, if reversed in its direction, hold these two forces in 
equilibrium, the above method serves also to determine in all 
respects the force capable of equilibrating the forces P and 
P x ; and it may be observed generally that if a force R holds 
in equilibrium any number of forces, it will, if reversed in its 
direction, become the resultant of these forces. 

This suggestion will enable us to find the resultant of two 
parallel forces Pand P x (Fig. 19) applied 
at A and B, and acting in parallel but 
opposite directions. For it is plain that 
if we determine the position of a point C 
upon the line A B prolonged, at which 
a force R will hold in equilibrium the '' FlG 
given forces P and P xi this force, when 
reversed in its direction, will become the resultant of P and P x . 

But when there is equilibrium, we have, from the preceding 

article, 

P=P 1 + R, or R = P-P l = R 1 , 

the resultant, which is therefore equal to the difference of the 
given forces. 

Likewise [33] 

R:P::AB:AC y 

PXAB PxAB 



,J» 










* R , 


A 


B 


C 




'?, 




"R 



or AC = 



R ~ P-P 1 



Since the general rule of the last article is applicable in the 
present case to the forces P, P„ and R, it also applies, since 
R = R v to P and P x and their resultant R x . 

[35.] If the two forces P and P x become equal to each 
other, then 

R 1 = P-P x =o, 

and AC — 5 = 00 . 



44 ELEMENTS OF MECHANICS. 

Hence we conclude that no single force can equilibrate two 
equal and opposite parallel forces applied at any two points. 

It will appear also from the above that the less R i is, or the 
difference between the given forces, the greater is the distance 
of its point of application, and when P = P 1 this distance be- 
comes infinite. That there can be but one force which, taken 
in connection with P v will equilibrate the force P, and hence 
only one resultant to P and P v will appear by considering that 
the value of this force equal to R = P — P 1 is fixed, and the 
same force could not balance P and P x when applied at differ- 
ent points of the line A C. 

[36.] If there be any number of parallel forces in the same 
plane, and we determine according to the above method the 
resultant R of any two of them, this resultant may in turn be 
combined with a third force of the system, producing a second 
resultant, R r This again may be combined with a third force 
of the system, giving a resultant R 2 , and the process continued 
until all of the original forces have been employed and afnal 
resultant obtained for the whole system. 

OF MOMENTS. 

[37.] The product of the number representing any force 
by the perpendicular distance of its line of action from any 
point is called the moment of the force with respect to the point. 

If we conceive an axis to be drawn through the point per- 
pendicular to the plane containing the point and the direction 
of the given force, this product is called the moment of the force 
about the axis. 

If a force, making any angle in its line of direction with a 
given axis, be resolved into two components, one parallel to 
the axis, and one perpendicular to it in direction, the product 
of the latter component into the perpendicular distance of its 
line of direction from the given axis, becomes the moment of 
the force about the axis ; the parallel component evidently pro- 
ducing no tendency to rotation about the axis. 



ST A TICS. 45 

It would be well for the student to construct illustrative 
diagrams for the three cases referred to. 

The point with reference to which moments are taken is 
called the centre of moments. 

[38.] The moment of a force about any given point used 
as a centre of moments is the measure of the tendency to pro- 
duce rotation about the point, as may be shown thus : Let 
the fixed body B (Fig. 20), placed at 
the unit's distance from 0, a centre of *^± — a * x 



moments, be supposed to balance sepa- 
rately the forces P and P x , and denote by 
/ smdp x the pressures produced by them 
respectively at the point A . Since both 

of these pressures are produced at the same distance from 
the centre of moments O, they will plainly be to each other 
as the rotary tendencies of P and P x . But [33] we have 

p : P :: Ox : 1, 

p x :P x :: Ox x -.i; 

whence p : p x :: / \ O x : P x . x x ; 

or the moments of P and P x are as stated above. Assuming 
P x to be a unit force, and O x x the unit distance, P.O x will 
denote the measure of the moment of P, and so for any other 
force. 

[39.] Since when a given force acts at any point of a plane 
or solid, whose particles are rigidly connected, we may always 
conceive a line of particles, rigidly connected, lying in the 
direction of the applied force, it is obvious that its point of 
application may be changed to any point on this line without 
change of mechanical effect, since 7 the action of the force 
really extends throughout the line at whatever point it may 
be applied. 

[40.] Let R (Fig. 21) be any force applied at a points, 
having the point C as a centre of moments. Resolve this 
force into the two components / and P, the one coinciding in 



46 ELEMENTS OF MECHANICS. 

direction with the line joining the centre of moments and its 
point of application, and the other perpendicular to it. Draw 
C B = r perpendicular to A B, the line 
of direction of the given force R. Then 
will Rrbe the moment of the force R 
with respect to C as a centre of moments. 
But similar triangles ABC and A m n 
give 

R : A C = p :: P: r, 

or Rr — Pp. Fig. 21. 

Hence Pp, or the moment of the component perpendicular to the 
line joining the centre of moments with the point of application of 
the given force, is equal to the moment of this force. 




Static Couples. 



[41.] The combination of two equal and opposite parallel 

forces applied at the extremities of a straight line, as seen in 

[35], when the resultant R x becomes equal to zero, is called a 

Static Couple. The effect of a couple is to produce rotation 

about the centre of the line joining the points of application 

of the forces, without any motion of translation. The moment 

of a couple, being equal to the sum of the products of each 

force by half the distance between their points of application, 

and the forces being equal, is equal also to the force of the 

couple, multiplied by the whole distance between the points 

of application. A couple whose forces are Pand P l} and the 

distance of the points of application apart equals a line A B, 

P 
is represented by the symbol -p A B, which of course must not 

be understood to be an ordinary algebraic expression. 

PROPOSITIONS RELATING TO COUPLES. 

[42.] A couple may be revolved in its own plane without 
change of mechanical effect. 



ST A TICS. 



47 




Let - a b (Fig. 22) be the revolved position of the couple 
A 
p 
-5- A B, and apply at a and 

b the forces /„ and /, equal J 
and opposite to the forces/-*-- 
and /j. These four forces, 
mutually destroying each 
other, have no effect upon 
the original couple. 

Compounding the forces P and /„ and also P l and p 3 , whose 
points of application [39] may be transferred to their intersec- 
tions at and o v there will be produced two resultants R and 
R v equal and opposite to each other, and acting through the 
centre of moments C. They will therefore cancel each other, 

p 
and the couple — a b alone remains as the equivalent of the 

original couple. 

[4-3.] A eouple may be transferred to any other portion of its 

own plane without change of effect. 

P 
Let -pAB (Fig. 23) be the couple to be transferred. Take 

P p 
ab parallel to A B and form the equilibrated couples ab. 

A A 
These, mutually destroying each other, are of no effect. Join 

A b and Ba, bisecting each other at O. The forces Pand p s 
give a resultant R = P-\-p 3 , and the 
forces P x and /„ the equal and opposite 
resultant R x — P 1 -f- A- These cancel- 
ling each other, there remains the 



couple - a 
A 



as the equivalent of the 

p 

original couple -~AB. The couple 



A 


1 


K 


r 




\ 






3 




a 










'*. 








% 




■ 


A 



L 



Fig. 23. 



# <5 may now be revolved into any position in its own plane. 



Hence the truth of the proposition : 



4 8 



ELEMENTS OF MECHANICS. 




Fig. 24. 



[44.] A couple may be transferred to any parallel plane in a 
mass without change of effect. 

P 
Let -p AB (Fig. 24) be the given couple lying in the plane 

M N. Take M X N V a parallel plane, and draw the equilibrated 

couple ? a b parallel to -5- A B. This will not disturb the 

A A -m 

effect of the original couple, nor 

have any effect upon the mass in 
which the planes are supposed 
to lie. Draw the lines A b and 
Ba, bisecting each other at the 
point O. The forces P and p 3 
compounded give a resultant 
R = 2P, and the forces P 1 and 
p 2 the equal and opposite result- 
ant R x . Hence there is left, as 
the only effective force of the 

p . P 

system-, the couple -— a b as the equivalent of -5- A B, which 

may also be revolved [42], if necessary, in the plane M X N V 

[45.] It follows, from what precedes, that static couples ; 
having equal moments and lying in the same or parallel 
planes, are equivalent ; and hence one couple may be repre- 
sented by another, even where the distance between the 
points of application is different, provided only the equality 
of the moments of the two couples be preserved. 

If it be required to transfer a couple, so that it may pass 
through a given point, we have onlp to pass a plane through 
the point parallel to the plane of the couple, transfer the 
couple to this plane, and then in the auxiliary plane to the 
required point. 

The distance apart of the points of application of any 
couple may be changed, provided the force of the couple 
is so changed in value as to preserve an equality of 
moments. 



STA TICS. 



49 



COMPOSITION OF ROTATIONS. 



P P 

[46.] Let -pAB and -— AB (Fig. 25) be two static couples 

tending to produce a revolution of the material particles A 
and B rigidly connected in the circles A D E and A FG, whose 
axes pass through the centre of moments C, and are respec- 
tively perpendicular to the planes of the circles. 

The particles at A and B being acted upon by the two 
forces P and P jy tangents to the cir- 
cles of revolution, will tend to move 
in the direction of the equal result- 
ants P and R v which together con- 
stitute a resultant couple tending to 
produce a revolution of the par- 
ticles in a circle lying between the 
former two. In like manner a third 
couple might be compounded with 

the couple 77 AB, giving rise to an- 

other resultant motion of rotation; 

and the same principle might be 

extended to any number of original couples. 

If we conceive two axes to be drawn through the centre 

of moments C coincident with the axis of revolution of the 

P P 

couples -pAB and ~ A B, they will evidently make the same 

angle with each other as the forces P and P 1 ; and since the 
resultant axis of rotation must also make the same angles with 
the axes of the given couples that the resultants of the forces 
of these couples make with these forces, it follows that if, 
upon the axes of the couples, lines be taken from C to repre- 
sent the forces P and P v and their resultant be found by com- 
pleting the parallelogram, this resultant will coincide with 
the resultant axis of rotation. 

The principle of the composition of rotations has an in- 




50 



ELEMENTS 0E MECHANICS. 



teresting application in the explanation of the astronomical 
fact of the precession of the equinoxes, as well as of the curi- 
ous mechanical contrivance known as the Gyroscope. 



Geometrical Constructions. 

[47.] The principles deduced in [33] and [34] may be 
conveniently applied to the geometrical construction of the 
various quantities involved in the discussion of parallel forces. 
A few examples will be given as illustrations. 

(1.) To constrtict the resultant of two parallel forces P and P x 
(Fig. 26) acting in the same direction. 

Take Bn = A P=P and AC=BP X = P V and draw Cn, 
cutting A B in m. 

Similar triangles will give 

Bn : AC :: Bm : Am, 



or 



P : P, 



Bm : Am. 



rf 711 









/ 


.. - 


' p 


R 





n 



Fig. 26. 



Hence [33] m is the point of ap- 
plication of the resultant. Drawing 
m R parallel to A P and equal to 
A P+ BP X = P-{- P lf it will obviously 
be the required resultant. 

(2.) Let P and P x be supposed to act in parallel but opposite 
directions (Fig. 27). 

Make AB = BP=P. Take BE = AP l = P v and through 
D and E draw D C, meeting A B pro 
duced in C, which, as will be shown 
is the point of application for the re 
sultant. 

Similar triangles give 

AD : BE :: AC: BC, 
or P:P X ::AC:BC, 

or P: (P-P,) :: AC: AB 

Hence P : R 
°f [33]; an d since EP = BP— BE =.P— P, = R, we have, 




Fig. 27. 
But I34] P-P l = R. 
A C : A B, conforming to the general rule 




ST A TICS. 5 I 

by completing the parallelogram PC, the line CR — EP= R 
for the resultant ; C being its point of application. 

(3.) To construct the parallel components of the force R (Fig. 
28) when they lie in the same direction and are applied at given 
points A and B. 

Take A D equal and parallel to CR — R. Draw DB, cut- 
ting CR in E, and EP parallel to 
A C Completing the parallelo- 
gram EP V the required compo- 
nents, as will be shown, will be 
AP^PandBP^P,. 

For we shall have P-\-P 1 — AP 
+ BP, = CE+ER = R, the first 
condition necessary [33] ; and from similar triangles, 

AP: PD :: BE : ED :: EC: AC, 
But AP=P and PD=ER = P X . 

Hence P : P x :: B C : A C, 

and, by the general rule of [33], C is the point of application 
of the resultant of P and P x , 

(4.) Let AP=-P be the force of which the components ap- 
plied at B and C are to be found. 

Referring to the last diagram, draw indefinite lines CR 
and BP X parallel to A P. Draw PE parallel to A B. Pro- 
duce BE to meet A P prolonged in D. Take ER = PD and 
complete the parallelogram E P x , and make BP i = BP 1 ; then 
will CR = R and BP^ — P 3 be the required components, for, 
from the preceding example, 

P\P X \\ BC\AC But P X = P* 

Hence P : P 2 :: B C : A C, 

conforming to the general rule for parallel forces [33]. 
Likewise, by construction, R = P-\-P x = P-\-P tJ and hence 
P— R — P 2 , as should be the case [34]. 

Examples for actual construction may be furnished by the 
teacher if deemed necessary. 



52 ELEMENTS OF MECHANICS. 

Of Parallel Forces not in the Same Plane. 

[48.] Having shown in [33] and [34] in what mariner the 
value and point of application of the resultant of two parallel 
forces may be determined, whether they act in the same or 
opposite directions, it will be easy to determine the value and 
point of application of the resultant of any number of such 
forces ; this point being also known as the centre of parallel 
forces, or of the system. 

Let P y P v P 2 , and P 3 (Fig. 29) be four given parallel forces 
applied at the points A,A V A 2 , and^4 3 situated in any manner in 
space, and conceive these points to 
be inflexibly connected by straight 
lines as shown in the diagram. Since 
P and P i act at the extremities of the 
line AA X , their resultant R and its 
point of application D may be deter- 
mined as in [33]. 

Drawing A S D, the forces R and 
P s , compounded in a similar manner, 
will give R x as their resultant, and B 
its point of application. Proceeding 

in like manner with respect to R 1 and the remaining force P 9t 
the resultant R 2 of the whole system, and C, its point of appli- 
cation, are determined. 

If any of the components should act in an opposite but 
parallel direction to the remaining forces, as, for example, P z , 
then the resultant of P 2 and R would be [34] R x = R — P z \ 
though its point of application would no longer be at the 
point B. 

In the first case, where the forces all act in the same sense, 
we should have 

K = K + P = R + P, + P, = P+P,+P, + P 3 , 

and in the second case, in which P % is supposed to be negative 
or to act in an opposite sense, 

X, = P+P, + P t -P,; 




ST A TICS. 



53 



and if any number of components should act in a negative 
sense, they would likewise become negative in the value of R^. 
Hence the resultant of a system of parallel forces is equal to the 
ALGEBRAIC SUM of the components, and may be expressed by the 
formula r> •%? P' 

It should be observed that in what precedes the compo- 
nent forces have not been supposed to take any particular 
direction, but simply to be parallel to each other. If the 
forces P and P x should change direction, preserving, how- 
ever, their parallelism, neither the value of their resultant 
nor its point of application would be changed ; and since the 
same is true for all the partial resultants of the system, it fol- 
lows that neither the resultant of the whole system nor its point 
of application will be affected by supposing the components to 
assume any parallel directions. 

(49.] If the points A, A lt A„ etc., be referred to three rect- 
angular axes, and we represent by 

x, y, z the co-ordinates of the point A, 



1\, *x 



x ii }'i 



a a 



A, 



etc. etc. etc., 

and by X, Y, Z the co-ordinates of the point of application of 
the resultant of the system, then, as will be shown, the values 
of these co-ordinates will become known, and consequently 
the point of application of the 
resultant of the system, when- 
ever the co-ordinates of the 
points of application and the 
values of the components are 
given. 

For let A and A x (Fig. 30) 
be the points of application of 
the given forces P and P v and 
B the point of application of 
their resultant. Denote the Fig. 30. 

abscissae of A, A v and B respectively by x y x v and X v 





z 






4 






B^^ 




-A 




A ^ 









1 






*f~ 




X 




17 




f 




I 


t 


/J 


L4 


1 





54 ELEMENTS OF MECHANICS. 

Draw AD parallel to EG, the projection of AA 1 on the 
plane of X Y, and Ex, FX., and Gx x parallel to the axis of F. 

The similar triangles A CB and ADA, will then give the 
proportion 

AC: CI? :: AB\BA X :: P x : P; [33] 

or, since AC—EF, and CD — FG> 

EF'.FG :: P t : P. 
But by reason of the parallels Ex, FX V and Gx v 

EF'.FG :: xX x : X t x, :: (X x -x)\ {x.-X^ 

Px + P.x, 



or 



P t \Pv. &-*) : (*, -^ and X, = y^ 



If we now regard (P + P,), the value of the resultant ap- 
plied at i?, as a single force to be taken in connection with 
P 2 applied at A„ whose abscissa is X^ we shall have, by a 
similar course of reasoning, regarding X 9 as the abscissa of 
the point of application of the resultant of (P-\- P x ) and P 2 , 
the equation 

{P+P x )X t ^P,x, 



'" (P+PJ + P, 



or, replacing (P-\- P,) X t by its value taken from the preceding 
equation, 

Px + A^ + PtXt 



X t = 



P+P + P 



and proceeding in a similar manner with the resultant force 
(P+Pi-\-P^) applied at X„ and a component P, applied at 
some point A„ we find for the abscissa of the point of applica- 
tion of the resultant of four forces 

Px + P,x t + P ,x, + P 3 x, 
A >~ P+Pi+ P i J r P 3 • 



STATICS. 55 

and generally, omitting the accent, 



X = 



P+Pi + P % + etc 



Since we could proceed in like manner in reference to the 
axes of Y and Z, there would plainly result the following 
analogous equations, in which Y and Z are supposed also to 
be co-ordinates of the point of application of the resultant of 
the system, viz. : 

V - / > + /> ^+^ + etc. 
/>+/>+/>+ etc ' 

_ />* + />*, + />*, + etc. 
*~ P + P i + P i + etc. ' 

in which y, y v z, z v etc., are co-ordinates of points of applica- 
tion of the component forces of the system. 

Since P-\- P 1 + P a + etc. = R, the last three equations may 
be written thus : 

RX= Px + P, x, -f P 9 'x t + etc. ; 
R Y= Py + P x y x + P,y, + etc. ; 
RZ= Pz + P x * x -\-P % * % + *to. 

In what precedes, all the components have been supposed 
to act in the same direction, or in the same sense. 

It will be easy to show, however, that if any of them act- 
ing in an opposite sense are considered as negative, and the 
second members of the equations above are supposed to rep- 
resent algebraic sums, they will continue to be applicable. 

Referring again to the last diagram, by a slight change of 
notation let B, A v and A be the respective points of applica- 
tion of P, P„ and their resultant R, where P and P g are sup- 
posed to act in opposite directions. Denote by x, x v and X 



56 ELEMENTS OF MECHANICS. 

the abscissae of B, A v and A respectively. Then, from the 
general rule of [33] as applied to [34], we have 

P:P, :: AA X : A B :: E G : EE:: {x x -'X) : (x - X). 

Whence Px-PX= P x x x - P, X, 

or (P-P l )X=Px-P 1 x 1 ; 

or, since [34] P— P x — R, 

RX=Px-P x x x . 

If a third force P 2 should now be combined with R, pro- 
ducing a resultant R v we should in like manner have, denot- 
ing by X x the abscissa of the point of application of R v and by 
x 2 that of P 2 , 

R 1 X J =RX±P 3 x,=Px-P l x l ±P i x„ 

in which the plus-or-minus sign before the last term will be 
used according to the direction in which P^ is supposed to act. 
If P % is positive in direction, we shall also have 





*, = />—/»+/>. 


But if negative, 






R.r=P-P x - />„. 



It appears, therefore, that when the algebraic sums of the 
second members of the equations, giving the values of X, Y, Z, 
and R, are supposed to be taken, these equations become gen- 
eral, and are applicable to any system of parallel forces, in 
whatever sense the parallel components may act. These 
equations should therefore be written in the general forms 



R = 


2P; 




2Px 2Px 

2P ~~ R ' 


or 


RX=2Px; 


2Py ?Py 
* ~ 2P ~ R ' 


or 


RY=2Py; 


2Pz 2Pz 


or 


RZ=2Pz. 



2P R 



STATICS. 57 

[50.] The product of a force by the perpendicular distance 
of its point of application from either of the co-ordinate planes, 
is called the moment of the force with respect to the plane. Thus 
Px is the moment of the force P with regard to the plane of 
ZY, and RXXhe moment of the resultant of the system with 
reference to the same plane. 

The last three equations of the preceding article establish, 
therefore, the truth of the following proposition, viz. : The 
moment of the resultant of a system of parallel forces with re- 
spect to any plane whatever, is equal to the algebraic sum of the 
moments of the components with respect to the same plane. 

[51.] If the moments be taken with reference to a plane 
YZ passing through the point of application of the result- 
ant, called the centre of the system, then X — o, and hence 
R X = 2 Px = o ; or, the algebraic sum of the mome?its of the 
components with respect to this plane will be equal to zero. 

Conversely, if the algebraic sum of the moments of the 

components with respect to any plane be equal to zero, and 

2Px o 
R be not equal to zero, then X = — 75— = -^ — o, and the centre 

of moments will lie in the given plane. 

[52.] If we suppose the points of application of all the 
components to lie in the same plane, say the plane of X Y, 
then we shall have z = o, z 1 = o, z 2 = o, etc. etc. ; and hence, 
if R be not zero, we have Z= o, and the position of the centre 
of the system would be determined by the two equations 

Xt= ~R~ and Y =^T' 

[53.] If we further suppose the points of application of 
the components to lie in the same straight line assumed to 
coincide with the axis of X, we have 

z — o, z 1 = o, etc., and y = o, y t = o, etc., 
and hence Z=o and Y=o. 



58 ELEMENTS OF MECHANICS. 

The value of X, the abscissa of the centre of the system, 
will then be determined by the equation 



Conditions of Equilibrium of the System. 

[54.] If we conceive all the parallel components acting in 
the positive direction to have a resultant R r applied at some 
point A v and those acting in the negative direction a result- 
ant — R^ applied at A v it is obvious that when there is equi- 
librium in the system, we must have, first, the partial re- 
sultants R x and — R t numerically equal, and, secondly, their 
points of application A 1 and A 2 coincident. 

Regarding R^ as essentially negative, the first condition 
would be expressed by the equation 

R 1 + R, = 2P=P+P l + P t + etc. = o. 

To express the second condition, let X, Y, Z represent the 
co-ordinates of the coincident points A l and A^ 

Denote by N the sum of the negative moments with respect 
to the plane of Z Y, and by M the sum of the positive moments. 
We shall then have [49] 

R t X=M and R,X=N. 
Hence (R, -\-R 2 )X = M+N= 2Px ; 

or, since R l -\-R i — o, 

2 Px = P x + P x x x + P> x 3 + etc. = o ; 
and similarly, for the planes of XZ and X Y, 

2Py = Py + P lfl + P^ + etc. = o 
and 2 t Pz — Pz + P x z x + P, z 2 + etc. = o. 

These equations, since they are based upon the assumption 
that the points of application A l and A^ of the equal positive 



ST A TICS. 59, 

and negative partial resultants are coincident, are further 
equations of condition necessary in case of equilibrium. 

If the points A x and A 2 be not coincident, and R x = R 2 
numerically, the second conditions would not be fulfilled ; and 
since the two partial resultants R x and R 3 would obviously 
form a static couple, equilibrium would not ensue. 

[55.] When the second of the conditions above, alone, is 
fulfilled, or when the last three equations obtain, the system 
will have a resultant whose numerical value is equal to the 
difference between R x and R 2 ; but since the algebraic sum of 
the moments with respect to each of the co-ordinate planes 
is equal to zero, the centre of the system [51] will lie in each 
plane, and hence be found necessarily at the origin of co- 
ordinates. 

If, in addition to this, the plane of Z Y be assumed to be 
perpendicular to the direction of the forces, the planes X Y 
and X Z will be parallel to the system, and hence to its result- 
ant, which, since it passes through the origin, will therefore 
coincide with the axis of X. 

If the origin be a fixed point, or if there be a fixed point 
in the axis of X, the resultant will be equilibrated by its re- 
sistance, and equilibrium produced. 

Hence, when parallel forces having a resultant are applied to a 
system of material points, rigidly connected, one of zvhich is fixed, 
and the sum of the moments of the forces with reference to two 
planes drawn through this point, perpendicular to each other and 
parallel to the system, is equal to zero, there will be equilibrium. 



Of Oblique Non-Concurring Forces in the Same Plane. 

DETERMINATION OF THE RESULTANT OF ANY NUMBER 
OF SUCH FORCES. 

[56.] Graphical Solution. — Let A, A lt and A^ (Fig. 31) 
be the points of application of three forces P, P v and P % . 

If we suppose the two forces P and P l to change their 



<5o 



ELEMENTS OF MECHANICS. 



points of application in the plane to [39], they will give a 
resultant R, applied at the same 
point. Prolonging also this result- 
ant and the force P 2 to meet at O x in 
their lines of direction, they give 
the resultant R x as the equivalent 
of the three given forces. 

The same method being appli- 
cable to any number of component 
forces, it is obvious that by con- 
tinuing the process indicated the 
resultant of any system of oblique forces in the same plane 
may be easily determined. 

[57.] Second Method. — By referring the forces to a system 
of rectangular axes. 

Let P, P x , P v etc. (Fig. 32), be any number of forces ap- 
plied at the points A, A v A„ 




Fig. 31. 



etc., and let a, a v <* 2 and /?, 
/?„ /? 2 , etc., denote respectively 
the angles they make with the 
axes of X and F 

Resolving P into its two 
components A B and A C, par- 
allel to the axes of Jf and F, 
we have 

A B = P cos a 

and A C = P cos fi. 



N 



N 



Na 



p<- 



z 



ot 



Fig. 32. 



If the remaining forces be 
in like manner resolved into their components, parallel to the 
axes, we shall have two groups of parallel forces ; viz., Pcos a, 
P x cos a lf P^ cos « 2 , etc., parallel to the axis of X, and P cos fi, 
P x cos /?,, P 9 cos /? 2 , etc., parallel to the axis of F. 

Denoting the resultant of the first group by X, and that 
of the second by F, we have [24] 

X=2Pcosa and Y=2Pcosfi. 



STATICS. 6l 

These two partial resultants, lying in the same plane and 
being perpendicular to each other, would, if produced, meet 
in some point .0 (Fig. 33), and their result- 
ant, which would plainly be the resultant 
of the whole system, would be equal to 



R= VX* + Y 2 = V{2Pcosa) 2 -\-(2Pcos/3)\ 



Kt 

k y 



■> 



Fig. 33. 



If we denote the ordinate of the point K, 
the centre of the first group of components, parallel to the 
axis of X, by Y v and the general ordinate of the points of ap- 
plication of these components by y lt and in like manner the 
abscissa of the point K v the centre of the group of compo- 
nents parallel to the axis of Y by X % , and the general abscissa 
of these components by x„ we shall have from section [52],, 
for the values of Y x and X v which are the co-ordinates of 0, 
the centre of the whole system or point of application of its 
resultant, the expressions 

_ 2Pcosfix 2 _ 2 P cos a ft 

^2 — y » 1 — v ,9 

which values fix the position of the point of application of the re- 
sultant of the system. 

To determine also the direction which this resultant takes 
with reference to the axis, we have [24] 

X ZPcosa a 7 Y 2Pcos/3 
, cos a = -J? — 75 and cos b = -5- = j- 

K K K K 

The component forces being therefore given in all respects, the 
preceding equations make known the magnitude, point of applica- 
tion, and direction of the residtant of the system, which is like- 
wise determined in all respects. 

Special examples for solution may be given by the teacher. 



62 



ELEMENTS OF MECHANICS. 



Of Moments Referred to a Point. 

[58.] Before proceeding to establish the equations of con- 
dition in the case of equilibrium of a system of oblique forces 
in one plane, it becomes necessary to consider some of the 
principles of moments referred to a point [37]. 

Let A B — R (Fig. 34) be the resultant of the two forces P 
and P 1 applied at the point A and acting about C as a centre 
of moments. Supposing each of the q 7if g f A w 

forces R, P, and P 1 to be resolved into 
components perpendicular to C A and 
coincident with its line of direction, 
as in section [40], the perpendicular 
components will be respectively A 0, 
A m, and A n. By the same article, 
the moments of R, P, and P x will be 
equal to A O X A C, A m X A C, and 
A n X A C. But A m is obviously equal to n. Hence 




A m + A n — A m -\-On — A O, 
and Am X A C+ A nxAC=AOxAC; 

or, the sum of the moments of the components P and P 1 will be 
equal to the moment of the resulta?it R. If, therefore, r, p, and 
/>, denote respectively the perpendiculars let fall from the 
centre of moments upon the lines of direction of the forces R y 
P, and P v we shall have, by reference to the same section, 

Rr = Pp + P i p r 

If we suppose the centre of moments to lie within the angle 
of the two forces, as at C v we shall obviously have, by resolv- 
ing the forces in a similar manner, 

A O x = A n x — 1 n 1 — A n l — A m x ; 
AO x XAC x —An x xAC x —Am x xAC x \ 



ST A TICS. 63 

or, denoting the perpendiculars from the centre of moments 
as before, 

Rr = P x p x —Pt, 

the moments of the components having different signs, and 
tending to produce rotation in opposite directions about the 
point C v Regarding the moments in one direction as positive, 
and those in the opposite direction as negative, we therefore 
conclude, generally, that the moment of the resultant of two 
forces applied at a given point, with respect to any point within 
the plane of the forces, is equal to the algebraic sum of the moments 
of the components. 

If the forces P and P x do not concur and are not parallel, 
they may be applied at the point of intersection of their lines 
of direction [39] ; and if parallel, the point of application and 
direction of their resultant may be found by [33], or in fact 
by the last equation, which, solved with regard to r, gives 

Pp + P£ Pp + Pj x 

R ~ P+P x 

equal the distance from O, measured on a line perpendicular 
to the given forces. 

If there be two other forces, P 2 and P 3 , we should in like 
manner have 

Kr x = P,p 3 +P 3 p 3 . 

Treating the two resultants R and R x by the same method 

gives 

R,r, = Rr+R,r x ; 

or, by substituting for R r and R x r x their values above, 

R % r, = Pp + P x p x + P,p 2 + P 3 p 9 . 

[59.] Regarding the direction of the motion of the hands 
of a watch as positive, and the opposite direction as negative, 
the force P x would, in case the centre of moments C x were 
situated within the angle P x A P, tend to revolve the plane in 



64 ELEMENTS OF MECHANICS. 

the negative direction, and the above equation would have 
been in that case, omitting the accent, 

whence we conclude that the moment of the resultant y with 
respect to a given point, assumed as a centre of moments,, is equal 
to the algebraic sum of the moments of the several components with 
respect to the same point. 

[60.] By reference again to Fig. 32, the moment of the 
component A B = P cos a, with reference to the origin O as 
a centre of moments, is P cos a .y, and that of A C = P cos /?, 
in a negative direction, is — P cos /3 . x. Hence, by the pre- 
ceding article, we have, for the moment of the force P, the 
expression 

Pp — P cos a .y — P cos fi . x. 

Note. — The student may easily derive by a simple diagram, 
where x and y are the co-ordinates of the point of application 
of the force P, the equation p =y cos a — x cos /3, and hence 
the foregoing equation, independently of the method given. 

That this expression for the moment of a single force will 
apply to any force, in whatever angle its point of application 
may be, and whatever direction it may take, when the proper 
signs are attributed to the co-ordinates of its point of applica- 
tion, and to the cosines of the angles made with the axis, will 
become apparent only by reference to the diagram (Fig. 32). 

Thus, for example, the force P 2 in the third quadrant gives 
the two components P 9 cos <* 2 and P % cos #,, the one acting 
through the line O x M and tending to. produce rotation in the 
negative direction, and the other through O x N in the positive 
direction. But in this case we have cos a^ and cos y# a both 
positive, and ,r 2 and j/ 2 both negative. Hence for the force 
P 2 the general expression 

P cos a .y — P cos f3 . x 
would become 

— P^ cos or a . y 2 -f- P 9 cos /? 2 . x„ 



ST A TICS. 65 

in which expression, as we have just seen, the moments of the 
components are properly represented as to their signs. Let 
the student consider in like manner the action of the com- 
ponents of the three forces P zt P if and P 5 applied at the same 
point O v and the form the general expression above will 
assume, and finally suppose the point to lie in any of the 
four quadrants. It will be found that this expression for the 
moment of any force about the origin, will properly indicate 
the moment of each component as to its sign. 

If, therefore, we have any number of forces whose points 
of application are referred to rectangular axes passing through 
the centre of moments, we have, from the preceding article, 

Rr= 2Pj>= 2 P{cos a .y — cos . x\. 

The moment of the resultant of any number of oblique 
forces in one plane with respect to a given point, assumed as 
the origin of co-ordinates, will therefore become known when- 
ever we have given the co-ordinates of the points of applica- 
tion of the components, and the angles made by them with 
the given axes. 

Solving the preceding equation with regard to r, we find 
for the perpendicular distance from the origin, or centre of 
moments, to the line of direction of the resultant, 



r = 



2P\cosa .y — cosfi.x\ 2 P\ cos a .y — cos /?. x\ 

2 P 5 cos a . y — cos fi . x\ 
1/(2Pcos tf) 2 + (2 Pcos fi 



66 ELEMENTS OF MECHANICS. 



Conditions of Equilibrium of the System. 

[61.] When the system is in equilibrium we must evidently 
have the resultant equal to zero, or 



R z= VX 2 + Y 2 = V(2 P cos of -f (2 P cos /?) 2 = O, 
whence, also [59], Rr = 2 Pp = o. 

The squares of the quantities under the radical being 
essentially positive, we must therefore have, as the equations 
of condition in case of equilibrium, 

2 P cos a = P cos a -f- P l cos a 1 -f- P % cos a 2 -f- etc. = o, 
2 P cos ft = P cos fi + P x cos A + P, cos A + etc. = o, 

and [60] 

2P{cos ay - cos/5^f = 2Pp = Pp-\-P x p x -f P 2 / 2 + etc. = o. 

Before proceeding to interpret these equations, it should 
be observed that the signs of the moments Pp, P y p v etc., of 
the last equation might be known by the rule of [59] ; but if 
we recur to [60] we have, for the perpendicular distance / 
from the centre of moments to the line of direction of the 
force P, the equation 

p ■= y COS a — x COS f3. 

It has already been shown, however, in the preceding 
article, that if we regard the force P as essentially positive, 
the expression 

y cos a — x cos /?, 

when proper signs are given to x, y, cos a, and cos /?, will of 
itself indicate the sign of the moment of the force P. 

Hence the sign of/, determined by the sign of the second 
member of the last equation, will make known the sign of Pp 
for the moment of any force, whenever we have given the co- 
ordinates x and y of its point of application, and the angles its 
direction makes with the co-ordinate axes. 





u 


i= 


' / x 








*'--- ■ 


1 
1 

1 

1 ^^ 


' \ 


jjT"" J * 


«* j - 






\# 


"4 


* * 



^7^ zycs. 67 

[62.] Let us now consider the meaning of the above equa- 
tions of equilibrium. 

The first of them, viz., 

P cos a -j- P 1 cos a x -f- P 2 cos or 2 -|- etc. = o, 

expresses the condition that the algebraic sum of all the com- 
ponents in the plane of XY parallel to the axis of X shall be 
equal to zero; so that if we denote by X (Fig. 35) the resultant 
of all the components acting 
in one direction parallel to 
this axis, and by X^ that of 
the components acting in the 
opposite parallel direction, 
these two resultants must, 
as indicated by the above 
equation, be equal to each 
other. Fla 35 ' 

In like manner, the second of the equations of condition, 

P cos fi + P l cos f3 1 + P 2 cos /? 2 -f etc. = o, 

shows that there are also two equal and opposite parallel forces, 
Fand Y v partial resultants of the components parallel to the 
axis of F. 

Prolonging X and F, and X x and Y lf until they meet re- 
spectively in the points P and P v and supposing these forces 
to be transferred to these points, we should have, by composi- 
tion, R and R 1 as the resultant or effective forces of the system. 
But since X= X x and Y= Y lf the resultants R and R lf the 
diagonals of equal rectangles, are also equal. 

They are, moreover, opposite in their directions and par- 
allel, as may easily be seen by observing the parallelism and 
equality of the sides of the triangles R NP and P x M R x . 

Hence it appears that the first two of the equations neces- 
sary to a condition of equilibrium, allow two equal and oppo- 
site parallel forces, R and R„ applied at two points, P and P v 
in the plane of the co-ordinate axes, unless by accident these 
two points coincide. 



68 ELEMENTS OF MECHANICS. 

We may therefore conceive these two forces as forming 
a static couple [41], and hence there could be no fixed point 
in the system capable of holding them in equilibrium ; other- 
wise the resistance offered by it would constitute a resultant 
to the two forces, which [35] has been shown to be impossible. 

We conclude, therefore, that the first and second of the 
given equations of condition, do not prevent a motion of rota- 
tion, but at the same time do not allow a motion of transla- 
tion in the system. 

If, however, the last equation, 

2Pj> = o, 

be satisfied in connection with the first two, there will then be, 
as we shall show, no rotation, and the system will be in equili- 
brium. For if we suppose a line to be drawn through the 
centre of moments, and perpendicular to the two parallel re- 
sultants R and R x of the system, or to their lines of direction, 
and denote by r and r, respectively the distances of these re- 
sultants from the centre, their moments will be respectively 
equal to Rr and R x r v Whence, if the centre of moments lies 
without the parallels, we shall have 

2 Pp — Rr — R x r x — o; 

and if within, 

2Pp = Rr + R 1 r 1 = o, 

in which equations r and r 1 are positive. From the first, since 
R — R v we have 

R(r— r t ) — o, 

whence r — r 1 = o, and either r = r 1 or both r and r\ are equal 
to zero. 

If r — r v the two parallel resultants R and R, will act in 
the same line, equally distant from the centre of moments 
and in opposite directions ; but if r = o and r l = o, this line 
will pass through the centre of moments. It is obvious that 
in either case the two forces, being equal and directly opposed 
to each other, will induce equilibrium. From the second 



STA TICS. 



69 



equation we have in like manner r-\- r l = o, which gives r = o 
and r 1 = o, and the two resultants will approach and coincide 
in a direction passing through the centre of moments, but act- 
ing in opposite directions. The value r == — r l is not admis- 
sible, since the centre of moments was supposed to be within 
the parallels, and both resultants tend to produce motion in 
the same sense. 

We conclude, therefore, that the resultants R and R v equi- 
librating each other, can produce neither a motion of trans- 
lation nor a motion of rotation, and hence, when the three 
equations of condition are fulfilled, the system must be in 
equilibrium. 



Of Oblique Non-Concurring Forces in Space. 

[63.] Let AX, A Y, and A Z (Fig. 36) be three rectangu- 
lar co-ordinate axes to which the 
force P, applied at the point O in 
space, is referred. 

Resolving this force into its 
three components parallel to the 
axes [20], and denoting by a, fi, 
and y the angles it makes with 
the axes of X, Y, and Z respec- 
tively, we shall have, as hereto- 
fore, P cos or, P cos fi, and P cos y 
for the three components parallel 
to the axes. 

Since every other force could in like manner be resolved 
into similar components parallel to the axes, we have, for the 
algebraic sum of all the components parallel to the axes of 
X, Y, and Z respectively, the expressions 

P cos a -f P 1 cos a x -f- P 2 cos tf 2 + etc. = 2P cos a ; 
Pcos/3 + P i cos fi t + P 2 cos fi % + etc. = 2Pcosfi; 
P cos y + P x cos y x + P 2 cos y % + etc. = 2P cos y. 




Fig. 36. 



70 ELEMENTS OF MECHANICS. 

Or, denoting the resultants of the several groups of parallel 
forces by X, Y, and Z, we have 

X=2Pcosa, Y=2-P cos ft-, Z=2Pcosy. 

These three partial resultants, being perpendicular to each 
other, will give a resultant to the whole system, under the 
following conditions : 

(i.) When the three forces X, Y, and Z concur in one 
point, and constitute the three edges of a rectangular paral- 
lelopipedon, of which the diagonal will be the resultant ; and 

(2.) When two of them concur, or lying in the same plane 
may be prolonged to meet, and the third in its line of direc- 
tion intersects their resultant, or this resultant produced. 

In either case, however, the value of the resultant will be 
the same, and equal to the diagonal of the rectangular paral- 
lelopipedon of which the three partial resultants X, Y, and Z 
form the edges. 

Hence, calling the resultant R, we shall obviously have 

R= yx* + Y 2 + Z\ 

[64.] To find the co-ordinates of the point of application of 
the resultant. 

In the first place, let the three partial resultants X, Y, and 
Z meet in a common point, in conformity with the first of the 
above conditions. Taking the components of the original 
forces parallel to the axis of Y t which give the partial result- 
ant denoted by Y, we have [50] the moment of this resultant, 
with respect to the plane of Z Y, equal to the algebraic sum 
of the moments of its components. 

Denoting by X x the abscissa of the point of application of 
the resultant Y, we therefore have 

YX^SPcosp.x, 

2 P cos fi.x 2 P cos fl . x 

or X, — 77 == ^, „ ^ — • 

Y 2 P cos ft 



ST A TICS. 



n 



In like manner, if we denote by Y x and Z x the two remain- 
ing co-ordinates of the point of application of the partial re- 
sultants X, Y, and Z, we have, from the moments of X with 
respect to the co-ordinate planes XZ and X Y, the similar 

equations 

2P cos oL.y 2P cos a.y 



Y t = 



and 



X 2 P cos a 

2 P cos a . z 2 P cos ol . z 
X 2 P cos a 



Since the partial resultants are supposed to concur in one 
point, the co-ordinates X v Y lf Z x of this point are also the co- 
ordinates of the point, of application of the resultant of the 
whole system. 

Hence, whenever the co-ordinates of the points of appli- 
cation of the original forces, their intensities, and the angles 
they make with the axes are given, the second members of the 
equations above, representing the co-ordinates of the point of 
application of the resultant of the system, become known. 

Denoting by a, b, and c the angle the resultant R makes 
with the axes, its direction [24] will be given by the equations 



cos a 



X 



cos b — 



Y 
R' 



and 



cos c 



£ 

R 



In the second place, let the three partial resultants X, Y r 
and Z (Fig. 37) fulfil the second 
condition referred to in the pre- 
ceding article, and suppose X 
and Y to be prolonged to meet 
at the point 5 in space, whose 
projection upon the plane of 
X Y, parallel to the plane of 
the two forces X and F, is O x . 
Let P be the point at which the 
force Z produced would pierce 
the plane of X Y, and draw P0 1 
prolonged to Q. 

The lme PQ will then be that Fig. 37. 




72 ELEMENTS OE MECHANICS. 

in which the resultant of X and Y must be projected upon the 
plane X Y, in order that this resultant may meet the direction 
of the force Z produced, as premised in the second condition. 

The partial resultants X and Y being both parallel to the 
plane of X Y, their resultant will also be parallel to the same 
plane, and equidistant from it at all points. Hence, taking 
PO = vS V or equal to Z v the ordinate parallel to the axis of 
Z of the point of application of either of the forces X or Y, 
we have O as the point of intersection of the resultant of X 
and Y with the force Z produced, and hence, also, as the 
point of application of the resultant. To find the co-ordinates 
of this point we have therefore only to find the two co-ordinates 
X x and Y x of the point of application m of the partial resultant 
Z, and the one co-ordinate Z x of the point of application of either 
of the resultants X or Y, which, as seen by the diagram, are 
equal to the co-ordinates of 0, the point of application of the 
resultant R. 

We shall therefore have, for the required co-ordinates, 

Y _ 2 P cosy, x _2Pcosy.y _'2Pcosa.z 

1_ ^PcosTP l ~ ZPcosy ' x ~ SPcosa ' 

in which equations, as before, the second members involve 
only given quantities, and the direction of the resultant and 
its intensity will be determined as in the case above. 



Of Moments with Respect to an Axis. 

[65.] In order to determine the conditions of equilibrium 
of the system of forces treated of in the preceding sections, it 
becomes necessary to consider the principles of moments with 
respect to an axis. 

The moment of a force with respect to an axis is its tend- 
ency to produce rotation about the axis [37]. 

Recurring again to Fig. 36, let P be any force applied at 
the point and resolved into its three components, 

P cos a, Pcos/?, and Pcosy. 



STA TICS. 73 

The tendency of the force P to produce rotation about the 
axis of Z, will obviously be equivalent to the tendency of its 
components to produce a like result. 

Of these components, P cos y, being parallel to the axis of 
Z, will be of no effect in producing rotation. The component 
P cos a acts with a potential arm equal to Z B, or to the co- 
ordinate y, to produce rotation in a negative direction, and the 
component P cos fi with an arm Z C, equal to x, to produce 
rotation in the positive direction. The whole tendency to 
rotation, or the moment of P with respect to the axis of Z, 
will therefore be given by the expression 

{Pcos /3 .x — Pcos a.y\. 

That this is a general expression for the measure of the 
tendency to rotation about the axis of Z, for a force applied 
at a point in any one of the eight triedral angles formed by 
the co-ordinate planes, will become apparent only by consider- 
ing the subject in a manner similar to that explained in [60], 
to which the student is referred. Conceiving the force to be 
applied successively at points in each of the eight angles, and 
to assume any direction, it will be found that when the proper 
signs are attributed to the co-ordinates of the point taken, and 
to cos /3 and cos a, the given expression will truly represent 
the moment of the given force, the angles being reckoned 
from AX and A Y, as in [24], or from lines parallel to these 
through the point of application. 

In like manner, since every force in the system would give 
a similar expression as a measure of its rotary effect about the 
axis of Z, and since, by parity of reasoning, like expressions 
could be deduced for the tendency to rotation about the axis 
of X and F, we have, by reference to [64], and denoting by 
X iy Y v Z x the co-ordinates of the points of application of the 
resultants of the groups of components, parallel to the axes, 
the general expressions 



74 ELEMENTS OF MECHANICS. 

2Pcos p. x~2P cos a. y = YX X - X Y x for the axis of Z» 
2Pcos y.y — 2Pcos/3.z = Z Y x - Y Z x " " " " X; 
2Pcos a.z— 2 Pcosy ,x = XZ, — ZX, " " " " Y; 

or FX, - X Y x — 2 \P(cos fi . x - cos a .y)}, 

ZY X — YZ X = 2 {P(cos y. y - cos ft.z)}, 
XZ X - ZZ, = 2 {P(cos a.z-cosy.x)\, 

for the moments of the whole system, with respect to each of 
the co-ordinate axes Z, X, and Y. 



Conditions of Equilibrium of the System. 

[66.] In order that a system of forces situated in any 
manner in space, and acting at points in any way rigidly con- 
nected, or points of the same body, may be in equilibrium, 
the resultant action of the forces composing it must neither 
tend to produce a motion of translation, nor a motion of rota- 
tion about an axis. 

If we suppose the resultant of the system to be equal to 
zero, we shall have 



R = VX 2 + Y 2 + Z 2 = o. 
Whence X— 2P cos a — o, Y= 2P cos ft = o, 

and Z=2Pcosy = o. 

The equation 2 P cos a = o expresses the condition that 
the resultant of all the components of the original forces par- 
allel to the axes of X in a positive direction, is numerically 
equal to the resultant of those acting in a negative direction ; 
but as these resultants do not necessarily, and would not 
ordinarily, meet in a common point, or act in the same line 
[62], they would not generally destroy each other, but con- 
stitute a static couple. Similar remarks apply to the two 
remaining equations. 

Hence there would be formed three static couples, which 



ST A TICS. 7$ 

we may, without change of effect [45], conceive to be trans- 
ferred to a common centre of moments. 

They would thus tend to produce a rotation of the sys- 
tem about three different axes passing through this centre, 
the resultant effect of which would be a tendency to rotation 
about some one particular axis [46], whilst there could be no 
tendency to a motion of translation of the system. 

The above condition, therefore, that R — o is sufficient to 
prevent all motion of translation, but not of rotation. 

Again, if we conceive the rotary motion due to the effect 
of the three couples, the equivalents of the original forces, to 

be the effect of a single resultant couple -^ A B [46], the forces 

R and R x may be considered as the resultants of the system. 

If now R and R l be prolonged in their directions to meet 
either of the co-ordinate planes, assumed as planes of refer- 
ence, their moments with respect to one or both of the axes lying 
in this plane must, as we shall show, be unequal. For if the 
points of intersection are on the same side of each axis, they 
will necessarily be unequally distant from one if ?wt from both of 
them, and hence the moments of R and R 1 with respect to one 
if not both of the axes will be unequal, and a rotary motion 
will ensue ; and if they lie on opposite sides of either axis, the 
moments with respect to this axis will both have the same 
sign, or the forces act in the same sense, and still induce rotary 
motion. 

Hence, so long as there is a tendency to rotation, due to 
a resultant couple, about any axis whatever, there will also be 
a tendency to rotation about some one of the co-ordinate axes. 

If, therefore, we impose the further condition that there 
shall be no tendency to rotation about either of the co-ordinate 
axes, there can be no rotation in the system about any axis 
whatever, and equilibrium must ensue. 

But that there shall be no rotation about either axis, the 
algebraic sum of the moments with respect to each must be 
equal to zero ; that is to say, we must have, for the remaining 
equations of condition [65], 



j6 ELEMENTS OF MECHANICS. 

2 \P(cosj3.x — cosa.y)\ = o; 
2 {P(cosy .y — cos/3 .*)] =o; 
2 {P(cos a. z — cos y .x)\ = o. 

These conditions, being fulfilled, would cause the resultants 
R and R 1 of the supposed couple tt -A B to approach and 

meet in a common point of application, whereby they would 
equilibrate or destroy each other. 

[67.] Should the original forces concur at one point, the 
values of x, y, and z, for the points of application of these 
forces, will become respectively equal ; and if the first equa- 
tions of condition, 2 P cos a. = o, 2 P cos ft == o, and 
2Pcosy — o, be fulfilled, the last three will be also, and 
there will be equilibrium. See [31]. 

[68.] The foregoing equations of condition, having refer- 
ence to forces situated in any manner in space, are therefore 
general in their application, and we should be able to derive 
from them the equations of condition in all the cases of equi- 
librium treated of in preceding sections. 

If, for example, we suppose the forces of the system to be 
confined to one plane, as that of X Z, the angles formed by 
each force with the axis of Y would be equal to 90 , and the 
co-ordinate y of its point of application equal to zero. We 
should therefore have cos fi = cos 90 = o, and y = o, and the 
general equations of condition, for any system of forces, would 
become, under these restrictions, 

2P cos a — o, 2P cos y — o, 
and 2 \P(cos a .z — cos y . x)\ —0\ 

which equations, if it be simply observed that the axes of Y 
and Z have been interchanged in the discussion, are found to 
coincide with those of [61]. 



STA TICS. 77 

[69.] Should the forces concur at one point, the first two 
equations, being fulfilled, would also satisfy the last equation, 
and they would therefore be the only necessary conditions in 
case of equilibrium. Hence, if any number of forces in the 
plane of X Y concur at the origin of co-ordinates, there will 
be equilibrium when 

2P cos a = o and 2P cos ft = o. [24] 

[70.] If we suppose the forces to be a system of parallel 
forces situated in any manner in space, the values of cos or, 
cos ft, and cos y for any assumed system will be constant 
quantities, and the general equations of [66] will become 

cos a 2P =0, cos ft 2P = o, cos y 2P = o, 

and cos ft 2P.x — cos a 2P.y = 0, 

cos y 2P.y — cos ft 2 P. z = o, 
cos a 2 P. z — cos y 2P.x = o. 

But from the first set of equations, since cos a, cos ft, and 
cos y are not equal to zero, we have 

2P=o. 

Again, take the first of the second set of equations, and 
observe that if the forces make the angles a and ft with the 
axes of X and Y (or, which is the same, with lines drawn 
through their points of application parallel to these axes), it 
is possible for a to become less whilst ft becomes greater, or 
vice versa, whilst at the same time neither the co-ordinates of 
the points of application nor the magnitudes of the forces are 
affected, and 2 P. x and ~2 P .y remain constant. 

But it is evidently impossible for this equation to be satis- 
fied whilst cos a and cos ft may change value in the manner 
referred to, and 2Px and 2Py remain constant and finite. 

Hence we must necessarily have 2 Px = o and 2 Py = o, 
and in like manner, from the second equation, 2 Pz = o. 



78 ELEMENTS OF MECHANICS. 

Under our supposition, therefore, the equations of conoi- 
tion become 

2P=o, 2Px = o y 2Py = o, 2Pz = 0; 

and these are the same equations as have already been deduced 

in [54]. 

[71.] Should the forces all lie in one plane, as ZZ, then 
P = 90 and y = 0, and the only necessary equations would be 

2P=o, 2Px = o, 2Pz = o. [52] 

If we further suppose the points of application to be in the 
axis of X, then 

z = o, 2P=o, and 2Px = o. [53] 

[72.] It appears from the preceding sections that the 
shorter method might have been to consider first the general 
case of forces applied at any points whatever in space, and, 
by imposing the necessary conditions for each specific case, to 
deduce the results which have been already given. The 
inductive process adopted, however, was deemed more suit- 
able for those just entering upon the study. 



Centre of Gravity. 



[73.] In every body or system of bodies, supposed to be 
rigidly connected and acted upon by gravity alone, there is a 
point which if supported would preserve the body or system 
in a state of equilibrium in any position of its component parts. 
This point is called the centre of gravity of the body or sys- 
tem. The lines of attraction upon the different particles of a 
mass converge toward the centre of the earth, but the dimen- 
sions of such masses as are treated of in Mechanics, in this 
connection, are so small in comparison with the distance to 
the earth's centre, that this convergence is too slight to give 
rise to any appreciable error in supposing them to be parallel. 



ST A TICS. 



79 



It may be observed that it would require a mass nearly 70 
miles in diameter at the earth's surface to produce i° of con- 
vergence between the outer lines of attraction. If we con- 
ceive, therefore, a body to be made up of an indefinite number 
of elementary particles, each of which is influenced by an 
equal attractive force, whose lines of direction are parallel to 
each other, the resultant of all these forces would be a single 
force passing through the centre of the parallel forces [49], 
which if reversed in its direction would hold the system in 
equilibrium. 

By the same article, also, the parallel components may take 
any direction without changing the magnitude or point of appli- 
cation of the resultant. 

But if we conceive the body or system to be revolved into 
a new position of its component parts, it will be simply equi- 
valent to changing the direction of the parallel components 
represented by gravity in their manner of acting upon the 
body or system. 

It follows, therefore, that the centre of the parallel compo- 
nents due to the force of gravity, or the point of application of 
their resultant, is likewise the centre of gravity of the body or 
system. 

The position of the centre of gravity of some bodies, which 
have their particles homogeneous throughout, and are sym- 
metrical in form, is very obvious, without the aid of a demon- 
stration; as, for example, the straight line, the circle, the 
sphere, the ellipsoid, etc. 

Before proceeding to investigate a general formula for 
finding the centre of gravity, several examples will be .given 
that are easily solved by special methods. 

EXAMPLES. 

[ 4.] To find the centre of gravity of a plane triangle ABC 
(Fig. 38). 

Draw the line A D from the vertex A to the middle of the 
side B C, and also B E to the middle of A C. It will be shown 




SO ELEMENTS OF MECHANICS. 

that the centre of gravity of the triangle is at 0, the intersec- 
tion of A D and B E. Conceiving 
the triangle to be divided into an 
indefinite number of elementary 
lines, m n, etc., parallel to B C, the 
centre of gravity of each one of 
these elements will be at its centre. 
But since A D bisects B C, it will FlG - 38. 

also bisect all the elements parallel to B C, and pass through 
the centre of gravity of each. Should we apply at each of 
these centres of gravity a vertical force equal to the weight 
of the corresponding element, every element, and conse- 
quently the whole triangle, would be held in equilibrium by 
a system of parallel forces applied at all points along the line 
A D, and diminishing in intensity from D to A as the elements 
decrease in weight. But since all these forces lie in a vertical 
plane passing through A D, their resultant must act at some 
point on the line A D, which point will manifestly be the 
centre of gravity of the triangle. By parity of reasoning it 
may be shown that the centre of gravity will also lie on the 
line BE, bisecting the side A C and the elements drawn par- 
allel to it. Hence it must be found at O, the intersection of 
these two lines. 

As a geometrical problem, let the student show that 
EO — \EB,oyDO=.\DA. In subsequent portions of the 
text it will be assumed that this has been done. 

If we suppose three equal particles to be placed at the 
points A, B, and C, the two particles B and C might be brought 
together at their centre of gravity Z>, when O would plainly 
be the centre of gravity between these two and the particle 
at A. Hence the centre of gravity of three equal particles at 
the vertices of a triangle is the same in position as the centre 
of gravity of the triangle itself. 

[75.] To find the centre of gravity of any plane rectilineal 
figure A B C D E (Fig. 39). 

Dividing the figure into triangles, as shown in the dia- 



STA TICS. 



81 




Fig. 3g. 



gram, and finding the centres of gravity of these triangles by 
the method already given, join C C lt connect- g 

ing the centres of gravity of the first two. 
The triangles ABC and A C D may be con- 
ceived to be sustained against the action of 
gravity by the application of two parallel ver- 
tical forces equal to their respective weights 
at the points C and C 1} their centres of gravity. 

The point of appplication of the result- 
ant of these two forces will be the centre of gravity of the 
quadrilateral A B C D. 

To find this point, C 2 , the line C C x [33] must be divided 
into two parts, C C, and C x C„ which are inversely as the forces 
applied at C and C iy or inversely as the weights or areas of 
the triangles A B C and A CD. Join C 2 and C z , the centre of 
gravity of A DP, and divide also the line C t C % at the point C 4 
into parts C 2 C A and C s C i inversely as the areas A B C D and 
AED, and C K will be the centre of gravity of the whole figure. 
We have in the above, as will be observed, proceeded upon 
the hypothesis that the figure possessed weight, and that the 
weights of any portions of it were proportional to the areas 
of those portions. It is plain that the same method is appli- 
cable to a figure of any number of sides. 



[76.] To find experimentally the centre of gravity of a physi- 
cal plane of uniform thickness and homogeneous throughout. 

Let. the plane A B (Fig. 40) be suspended 
so as to move freely upon the axis CD, pierc- 
ing it at the point O, and from this point 
suspend a plumb-line OP across the figure. 

It is to be shown that when the plane 
ceases to vibrate on the axis CD, its centre 
of gravity will lie immediately under the 
line OP. 

For if not, suppose it to be at some point 
G. If we then suppose the weight of the 
plane to be concentrated at this point, and to be represented 




82 



ELEMENTS OF MECHANICS. 



by G W, it may be resolved into the two forces G F and G H. 
In this case the force G F will simply produce a pressure 
upon the point of support O, whilst G H will be a force caus- 
ing the point G to vibrate until it falls into the line O P, where 
the component GH will cease to exist. In like manner, if O x 
be taken as a point of support, we should find another line 
across the surface of the figure in which would be found its 
centre of gravity. The intersection of these two lines would 
therefore determine its position. 

[77.] To determine the position of the centre of gravity of a 
triangular pyramid (Fig. 41). 

Conceive the pyramid to be divided up into an indefinite 
number of elementary triangles, as I KL y 
parallel to the base B D H, and draw A G 
to the centre of gravity of the base found 
by the method already given [74]. 

It will be shown that the centre of 
gravity of each one of these elementary 
triangles which make up the pyramid will 
be found upon the line A G. Hence the 
centre of gravity of the pyramid will also 
be found upon this line [74]. 

Draw BE and A E to the middle of 
D H, the line A E cutting L K in its mid- 
dle point N. Draw also A G and IN. 

Since IN and B E are the intersections of the plane ABE 
with two parallel planes B H D and IKL> they are parallel 
to each other ; and since A G and IN both lie in the plane 
ABE, they will intersect at some point O in the plane I KL. 

Since N is the middle point of L K, the centre of gravity 
of IKL must lie on IN 

The similar triangles AGE and A O N give 
AE : AN:: EG: NO, 
and A EB and A N I give 

AE: AN:: EB: NI 
Hence EG: EB :: NO : NI 




Fig. 41. 



STATICS. 83 

But EG = iEB, and hence NO = iNI t and is the 
centre of gravity of the triangle I KL. Since the same can 
be shown for every elementary triangle composing the pyra- 
mid and parallel to IKL, the centre of gravity of the pyra- 
mid, as before stated, will be upon A G. In like manner it 
will lie upon the line B F, drawn to the centre of gravity of 
the triangular face A D H, 

But A G and B F, lying in the same plane ABE, intersect 
at some point C, which is therefore the C. of G. of the pyramid. 

As a geometrical problem, the student may show that A C 
is equal to f A G, or C G = -| A C, which will be henceforth 
assumed. 

Hence, to find the C. of G. of a pyramid, draw a line from 
the vertex to the C. of G. of the base, and take off f of that 
distance for the required point. 

It will be left for the student also to show that the same 
rule is applicable to a pyramid where the base is any polygon 
whatever. 

Assuming, first, a quadrangular pyramid, suppose it to be 
divided into two triangular pyramids, with triangular bases. 
Find the centre of gravity of the whole base ; draw a line 
from the vertex to this point, and show that this line does and 
should divide the line joining the centres of gravity of the 
pyramids, in the same proportion that it divides the line join- 
ing the centres of gravity of the triangular bases. Then from 
similar triangles the truth of the rule above may be shown, 
and the same extended to the case of any pyramid. 

Referring to the latter portion of [74], let the student sup- 
pose four equal particles to be placed at the four vertices of 
a triangular pyramid, and show that their C. of G. will be 
coincident with the C. of G. of the pyramid. 

[78.] The spherical zone. 

Conceive the zone to be cut by a series of equidistant 
planes parallel to its bases, and let d denote the distance 
between its bases. By the principles of geometry the elemen- 
tary zones thus formed will be equal. But if the secant planes 



84 ELEMENTS OF MECHANICS. 

arc indefinitely near each other, the sections will be merely 
the circumferences of circles, with the C. of G. of each upon 
the axis of the zone, these circumferences being of slightly 
varying breadth but of equal area, and hence of equal gravity. 
The C. of G. will therefore lie in the middle point of the axis 
joining the centres of the two bases of the zone. 

A spherical zone with one base is called a spherical calotte , 
hence if we have a spherical calotte of altitude d on a sphere 
whose radius is equal to R, the distance of its C. of G. from the 

centre of the sphere will be equal to [R — 



If the calotte becomes equal to a hemisphere, then d = A\ 
and the distance from the centre of the sphere to its C. of G.„ 
measured upon its axis, will be equal to 

2 

[79.] The spherical sector. 

Conceiving the sector to be divided into an indefinite 
number of equal pyramids, with their vertices at the centre of 
the sphere, their bases would constitute the base of the sector. 
The C. of G. of each pyramid would be at £ the radius, or f A\, 
from the centre of the sphere. These centres of gravity 
would therefore all lie in and constitute a spherical calotte 
with radius equal £ R. And since the weight of each equal 
pyramid may be supposed to be concentrated at its centre of 
gravity, the C. of G. of the sector must be coincident with the 
C. of G. of the calotte ; and since the radius of the calotte is 
equal to £ R, we have [78], for the distance of its centre of 
gravity from the centre of the sphere, 

2 

If the sector become a solid hemisphere, 
J d = i . £ R = f R y and X = £ R - | R = | R. 



ST A TICS. 



85 



[80.] To establish a general method of finding t J ic centre of 
gravity of a system of bodies referred to a set of rectangular 
axes. 

Let m, m v m %i etc. (Fig. 42), be any number of bodies situ- 
ated in space, and referred to the 
rectangular axes AX, AY, and 
A Z. Since the force of gravity 
upon the different masses acts in 
parallel lines, and the resultant of 
the gravitating influences upon 
them is equal to the sum of their 



weights, 



which denote bv W, we 







\ m ' 


t 


\1X' 


•^ 




X 


x, X 




/ 


1* 



Fig. 42. 



shall have [49], representing by X, 

Y, and Z the co-ordinates of the 

point of application of the resultant, 

and by x, y, z, x v j\, z^ etc., the co-ordinates of the centres of 

gravity of the various masses composing the system, the 

equations 



WX=2m 



r. 



WY=2mj>, 



and 



WZ= 2 ms\ 



or 



X = 



m x 



Y 



my 



IV 



and 



Z = 



m 



U 



in which X, Y, and Z are the co-ordinates of the point of 
application of the resultant of the system, and hence, also 
[73], of its centre of gravity. 

These formulas for X, Y, and Z are expressed in terms of 
the masses, as represented by their weights, and the co- 
ordinates of their centres of gravity, which are supposed to 
be given. 

To apply them to a continuous body, it may be conceived 
to be made up of an indefinite number of elementary parts, 
;;/, m v ///.,, etc.; and denoting the whole mass by AT, the second 
member of the equation above for the moment, with respect 
to the plane of Y Z, becomes 



MX— sum of moments of parts with respect to Y Z, 



86 ELEMENTS OF MECHANICS. 

sum of moments of parts 2 m x 



and X = 



M - M 



and like equations would obtain with respect to the other co- 
ordinate planes. 

If the whole mass be supposed to lie in one plane, as that 
of X Z, we have 

__ 2 my _ o _ 

and the two remaining equations suffice for determining the 
two co-ordinates of the centre of gravity ; that is, 

2mx JSmz 

x =-^r and z =-ir- 

If, furthermore, the body lies in a single plane, as that of 
XZ, and is symmetrical in form, the axis of X may be sup- 
posed to divide it symmetrically, and pass through its centre 
of gravity, whereby also 



2mz 



= r Tr.= = 9, 



M M 

and the only formula necessary for determining the distance 
of the centre of gravity from the origin, measured on the axis 
of X, will be 

2mx 

X= ' IT- 
EXAMPLES. 

(a) There are three bodies whose masses are denoted 
respectively by the numbers 6, 8, and 12, and the co-ordinates 
of whose centres of gravity are : for the first, x = 6, y — 4, 
z = 3 ; for the second, x x = 10, y x = 5, z x == 3; and for the 
third, x^ = 8, j 2 = 7, and z, = 6. It is required to find the co- 
ordinates of the C. of G. of the system, referred to the same 
origin. 



ST A TICS. 



&7 



(b) To fi?id the C. of G. of a circular arc ABC (Fig-. 43). 

Draw the chord A C, and from O, the centre of the circle, 
the radius B perpendicular to it, as the 
axis of X. The C. of G. of the arc will 
obviously lie on the line OB, which 
divides it symmetrically. Conceiving 
the arc to be divided into an indefi- 
nite number of elementary portions, as 
AE = m, E G = m iy G I = m v etc., they 
will be so small that they may be con- 
sidered as straight lines. Draw E D per- 
pendicular to A C, m to the middle of 
AE, and mx perpendicular to OB. We 
shall then have, by the preceding article, 
the distance from to the centre of grav- 
ity of the arc, measured on the axis of X, equal to 

2 m x vi x -f- 7//j x j -|- 7// 2 x„ -f- etc. 




Fig. 43 



X 



M 



Arc 



But from the similar triangles Omx and AD E we have, de- 
noting Ox by x, 

m : B : : A D : x, 

or mx = R. A D\ 

and in like manner m l x x — R.EF, 

m^x„ = R . G H\ 
whence, by addition, 

mx -f- m 1 x l -f- ^2-^2 + etc - = ^M & ~\~ E F-\- G H -\- etc.f; 

or, observing that the quantity contained in the parenthesis is 
equal to the chord A C which denote by C, we have 

2 m x = R . C, 

which substituted in the equation above for the value of X 
gives 



X = 



R.C 

Arc* 



88 



ELEMENTS OF MECHANICS. 



or, denoting the arc by A, 



X = 



RC 

A ' 



When the arc becomes equal to the semi-circumference we 

have 

C = 2 R and A = n R ; 



whence 



X = 



2 R" _ 2 R 
7tR ~ 7t 



.637^. 



(c) The circular sector A B G (Fig. 44). 

Conceiving the sector to be divided into an indefinite 
number of equal triangles similar to A B N, 
describe the concentric arc FE, with a radius 
A F equal to f R, where R is the radius of the 
sector. The centres of gravity of all the tri- 
angles thus formed will lie in the arc FE. 

Supposing the weights of the triangles to 
be concentrated at their centres of gravity, as 
in section [75], the problem is resolved into 
finding the point of application of the result- 
ant of a system of parallel and equal forces, 
the vertical force of gravity upon each triangle applied at 
every point on the arc FE. But as in the preceding example 
the force of gravity was not assumed to be of any absolute 
value, this point of application for the resultant will be the 
same as the C. of G. of the arc FE, which gives therefore, for 
the C. of G. of the sector, 




Fig. 44. 



x= 



AF X chord FE 
arc FE 



or, since in concentric circles, under the same angles, arcs and 
chords are to each other as their radii, we have 

AF=$R, chord FE = % chord BG, and arc FE- fare BG; 

whence, by substitution above, 

f RX chord BG 



X 



arc B G 



ST A TICS. 



8 9 



or, denoting the chord and arc by C and A respectively, 

R.C 



X=| 



A ' 



Whence it is seen that the C. of G. of the sector is at f the 
distance of the C. of G. of the corresponding arc from the 
centre of the circle. 

Should the sector become a semicircle, the chord will be 
equal to 2R and the arc equal to n R, whence 



X 



3* 



9.4248 



= .424^. 



(d) The circular segment BCD (Fig. 45). 

The moment of the sector ABB, with respect to the plane 
of YZ, will be equal to the sum of the 
moments of its parts ; that is to say, 
equal to the moment of the triangle 
A B D plus the moment of the seg- 
ment BCD. Calling the triangle T, 
the Sire A, the chord C, the sector S, 
and the segment S v we have by the 
preceding example, for the moment of 
the sector, the expression 




SX 



c 2 

o . 8 



RC 

A ' 



For the moment of the triangle we have 



TX= T.%AE= T.%Vr*-~. 



But 
Hence 



4 



T=i 



R*- —) = i c(r* 

' 4 /• 



■)=K(*-£J; 



9 o 



ELEMENTS OF MECHANICS. 



and for the moment of the segment, denoting by Xthe abscissa 
of its C. of G., the expression S x X. 

We have therefore, from the above, the equation of 
moments 



or 



s l x=s,$- i 



But from principles of Geometry, 

S : A :: TtR 2 : 2 7tR, 
Hence S x X = -^ C\ and 



or 



S = 



X=z 



RA 

2 

12 5, 



that is to say, also, that the cube with its three edges equal to 
the chord, is equivalent to a volume whose base is the seg- 
ment, and whose altitude is twelve times the abscissa of the 
C. of G. of the segment, from the centre of the circle as an 
origin. 

If the chord be supposed to become equal to the diameter, 
the segment will become a semicircle, 
and we shall have 



X = 



SR* 



3* 



= .424R, 



6ttR 2 
as seen before. 

(e) The semi-ellipse A B C(Fig. 46). 

From the centre of the ellipse O, 
with a radius O B equal to the semi- 
major axis, equal to A say, describe the 
semicircle EB D. 

Draw H K, LM y etc., as elements 
of the semicircle parallel to D E, and 
denote these elements by m, m v m„ etc., 
and the corresponding elements G I> 
N P t etc., of the ellipse by n, n v n„ etc. 




ST A TICS. 



9* 



Representing by x, x v x v etc., respectively the distances of 
these elements from O, or from the plane of Z Y, and by X 

and X x the abscisses of the centres of gravity of the semi-ellipse 
and semicircle, we have [80] 



and 



X- 



X, 



n x -f- n x x x -f- ^ 2 ^ 2 + etc. 

« + «! + « a + etc - 
mx-\- m x x x -f- m^ x^ -f- etc. 



77Z 4" **i H~ ^2 ~j~ e ^c. 
But from principles of An. Geometry we have 



m : « 

Mj : n x 

m^ : # a 

etc. etc. 



B 

A : B, or # = -j m, n x 



A m » 



Hence, by substitution, 



etc. etc. 



-. \m x + m 1 x x + ^ 2 x * + etc - 5 
X = 2j == J£i ; 

^{m-^-m i + m i + etc.] 

that is to say, the centres of gravity of the semicircle and .fc?;/zz- 
ellipse are coincident, and at a distance from their common 
centre O [80, <:] equal to .424 A. 

(f) A portion of a common parabola ABC (Fig. 47). 

From the origin A set off 
equal distances d in both direc- 
tions on the axis of Z. From 
the points of division draw 
lines parallel to the axis of X, 
intersecting the parabola in the ^ 
points n, n v n 2 , etc., and join- # 
ing these points by lines par- # 
allel to the axis of Z, form the & 
rectangles 1, 2, 3, 4, etc., which ^ 
may be considered as elements 
of the area. Let a and b be co- Fig. 47. 



z 






••-' 




B 


. 


s**' 




: / 




\ 


S 


'(* 












.' X. 


*, 


*3 


^ 


a, 






3 


4- 

71 
























^--J 


G 



92 ELEMENTS OF MECHANICS. 

ordinates of the point B, and x\, x\, x\, etc., the abscissae of 
the outer side of each rectangle from the origin. We shall 
then have the following 

Areas of Rectangles. 

ist 2dx v 

2d 4 d(x^ — x t ). 

3d 6d(x t —.xj. 

4th .......... 8 d(x 4 — x,). 

(n - i )th 2{ii — \)d (,!-„,_! — x n _ 2 ). 

nth 2ud(x n — x n _ x ). 

A bscissce &f Centres of Gravity. 

ist \x\. 

2d {* + *(*. - ^)i-i^¥4 

3d {**.+£(*,- *.)}.= *(*.'+.*.)• 

4th {x s + i(x< - * 8 )} = i (x< + x z ). 

in — i)th \x n _ 2 + £ (^.j — x n _ 2 ) I = J O u _! + -r n _ 2 ). 

»th j^_i + i (* n — x n _ x ) ) = J 0„ -f- #»_i)- 

Hence we have, representing the masses of the elementary 
rectangles by their areas, to which they are proportional, 

2 m x = dx? + 2d (x?. - x?) + 3 d (xf - x.;) + 4 a? (.r ; - O 

+ . . . (»— 'l).d(x»-i'7-*« %) + *£(*» — *W-l*)i 

or 

2mx = d\ — xf— * 3 '— * s a -;r 4 2 - . . . - x*-%—x n _{}+ndx*. 

But the general equation of the parabola is y~ = m x, and 
hence we have 

d 2 = mx v (2 dy = in x„ x (3 d)* — mx v etc.; 



ST A TICS. 93 

whence we get 






x n -{ = J^r , and * n a = a" ; 



which values, substituted in the equation for 2 m x\ give 

b 5 
But nd = b, or d b = - • 

n 

and summing the series within the parenthesis, by the alge- 
braic formula 



2 ni x = ba' 



> j 6(;/-i) 5 +i5(^-iy+io(;^-i) 3 -(^-i) [ 
m* n° \ 30 f * 



When 11 becomes infinite, 

or 2 m x — b a — —5 ; 

or, since £ 3 = ;;/ a, tn = - a -, 



and hence 2 m x = b a 9 — — - = - ba 9 . 

Denoting by X, as heretofore, the abscissa of the C. of G, 
of the parabola, we have 

MX— 2mx-^ba\ 



94 ELEMENTS OF MECHANICS. 

But since the mass M represents the area of the figure limited 
by B C, we have 

M = f a by 

which value, substituted above, gives 

X=ia, 
the result deduced by the methods of the Calculus. 

(g) Let the student endeavor to apply the general method 
illustrated in the examples given, to the case of the surface of 
a hemisphere, and derive the same value for X that was found 
by the special method adopted in [78]. 

The geometric truth that all zones cut from the surface of 
a sphere by a series of parallel equidistant planes are equiva- 
lent, will be necessary to the solution, which is also brief. 



Theorem of Pappus, or the Centrobaryc Method. 

[81.] This theorem, which usually takes the name of Gul- 
din, its supposed discoverer, seems to have been stated with 
clearness by Pappus, a Greek mathematician, as early as the 
fourth century ; and as many distinguished mathematicians 
who have alluded to its discovery, seem to give credit rather 
to the latter as its author, it is here given as the Theorem of 
Pappus, though it is barely possible it may have been redis- 
covered by the author whose name it ordinarily bears. Vide 
Mathematical Monthly for March 1859, Cambridge. 

It may be stated in the following words : 

The volume or surface generated respectively by the 
revolution of a plane area or line about a given axis in its own 
plane, when they lie wholly on one side of the axis, is equal to 
the superficies of the given area, or the length of the given 
line, multiplied by the circumference described by the centre 
of gravity of the area or line, as the case may be. 



ST A TICS. 



95 



If we suppose a plane area A (Fig. 48), lying in the plane of 
XZ, to be revolved around the axis of Z 
in such manner that all of its elementary 
particles shall move in parallel planes 
perpendicular to this axis, and in the 
circumferences of circles having their 
centres in the axis, it will generate an 
irregular solid of revolution, whose cross- 
section is the given area A. The rev- 
olution in a similar manner of the 
curved line A would generate 
face of revolution. 




z) 



a sur- 

Fig. 48. 

If we denote by m, m v m„ etc., the weights of the elemen- 
tary portions of the area or line, by x i x v x„ etc., their respec- 
tive distances from the axis of Z, and by M the mass, or 
weight of the given area or line, we have for the sum of the 
moments of these elements, with respect to the plane of Y Z, 
the expression 

mx -\- m 1 x\ -f- m 2 x 2 -f- etc. ; 

and if X denote the distance of the C. of G. of the area or line 
from the axis of Z, we have 



X 



2 m x mx -\-m 1 x 1 -\- ;/z 2 x 2 -|- etc. 



M 



M 



or 



M X = m x -f- m l x x -f- ;/z 2 x 2 -f- etc. 



But if A denote the area of the plane figure revolved, or the 
length of the line A, and a, a lt a„ etc., the elementary areas, 
or elementary portions of the line, we shall have 

M : A :: m : a, M : A : : m x : a v etc. etc., 

Ma Ma, 



or 



m = 



A ' 



in l 



etc.; 



whence, by substitution, clearing of fractions, and dividing 

by M, 

A X — a x -j- a x x x -\- a 2 x 2 + etc. 



9 6 



ELEMENTS OF MECHANICS. 



Multiplying both members of this equation by 2 n, we have 

2 7t X.A =2 nx .a-{-2 n x x . a x -\-2 n x^.a 2 -|-etc. 

But supposing the elements a, a lt a 2 to be indefinitely small,. 
2 7t x, 2 7t x v 2 7t „r 2 would represent the circumferences of the 
circles they describe in their revolutions, and 2 n x . a, 2 7tx x . a Jt 
2 7tx 2 . a„ etc., the solids or surfaces of revolution they gene- 
rate, as the case may be ; and since these aggregated will con- 
stitute the whole solid or surface generated by the whole 
area or line, which we may denote by S, we have 

A .2?tX=S; 

and since 2 7t X is the circumference described by the centre 
of gravity of the area or line in its revolution, this result con- 
forms to the enunciation of the theorem. 

By means of the preceding equation, if we know the con- 
tents of the solid, or area of the surface generated, and the 
area or length of the generatrix, the value of X, the distance 
of its centre of gravity from the axis of Z, becomes known. 
On the other hand, knowing the area or length of the genera- 
trix, and the distance of its C. of G. from the axis of revolu- 
tion, will enable us, by the same equation, to determine the 
solid or surface generated ; — the centrobaryc method of deter- 
mining solids and surfaces of revolution. 



EXAMPLES. 



(a) To find the C. of G. of the semi-circumference ZB D (Fig. 49).. 
We have in this case 



X = 



S 



27T A' 



Assuming the diameter Z D to be the axis 
of revolution, the surface generated will be Y) 
that of a sphere with radius R= C Z. So 
also A is the length of the generatrix, or 
semi-circumference, ZB D. 




ST A TICS. 



97 



Hence, substituting the values of 5 = 4 7tR 2 and A = n R as 
given by the Geometry, 

2R 
X= — = .6 37 R, 

as before determined [80, &]. 

(&) The circular sector CAEB, the triangle CAB, and the 
segment AEB (Fig. 50). 

Denoting the length of the chord A B 
by C, and the arc AB by A lf we have for 
the values of 5 and A, as given by the Geo- 
metry, Q 
S=27tR.C.iR = %7rR*C, y / 

and A=iA 1 R. ' 



Hence, by substitution, 




X = 



S 



RC 



2 n A 
as in [80, c\. 

For the triangle we have 



S = A.%.2tt CD, 
and, by substitution in the general formula, 

5 



X = 



2 tcA 



= *CjP, 



as has already been shown [74]. 
For the segment we have 

and, making the area of the segment equal A = S„ 
as seen in [80, d\ 



2 n S 1 12 5/ 



9 8 



ELEMENTS OF MECHANICS. 



(c) A portion ABC of a given parabola (Fig-. 51). 

Let A Z tangent at the vertex A be the axis of revolution. 
The volume generated by the revolu- 
tion, as determined by the Calculus, or 
otherwise, will be equal to f the cylin- 



der, having the same base and altitude ; 
or denoting B C by b, and A D by a, 







5 


= ±7ta 


'b, 


and, 


for the 


area 

A : 


of the 
= f ab. 


parabola, 


Hence 


X: 


S 

~ ir A 


-f*. 



J 



J&. 



Fig. 51. 
as seen in [80,/]. 

(d) The semi-ellipse ABC (Fig. 52). 

The oblate spheroid generated by the revolution of the 
semi-ellipse about its conjugate diameter 
is, by the Calculus, 



S = i 7tA\B, 

and the area of the semi-ellipse, An. Geo. 
or Calculus 

A =i7tA.B. 
Hence 

S 4A 



X = 



2 7t A " 3 7t 



.424 A, 



B 



Fig. 52. 



as in [80, e\. 

In the examples above, the solids and surfaces of revolu- 
lution, due to the motion of the generatrix about an axis, 
have been taken from other sources, and the centre of gravity 
in the case of each generatrix determined thereby ; but as, in 
the cases presented, these centres of gravity had already been 
determined in preceding sections, by the substitution of the 
known value of X in the equation 



27tA.X=S, 



STATICS. 99 

the value of S, the solid or surface of revolution, would de- 
pend only upon A, the area or length of the generatrix; and 
this being known, the value of 5 becomes known. 



Theorem of Leibnitz. 



[82.] We introduce in this connection an interesting as 
well as singular theorem, said to be due to Leibnitz. It may 
be enunciated in the following form : 

If a system of forces in equilibrium, acting in any direction in 
space, concur at one point, and are represented in direction and 
intensity by lines drawn from the point, this point will be the 
CENTRE OF GRAVITY of a system of equal particles, placed at the 
extremities of the lines representing the forces. 

Let P x , P„ P z , etc., be the given forces in equilibrium, and 
applied at the origin of a system of three rectangular co-ordi- 
nates X, Y, Z. 

Conceive any line A X to be drawn through the origin A 
to represent the axis of X, and each force to be resolved into 
its three components along the three axes. Denoting by a v 
a„ etc., the angles made respectively with the axis of X, we 
shall have, since by hypothesis the forces are in equilibrium, 

2 Pcos a = P x cos a x -\- P 2 cos a 2 -\- etc. == o. [31] 

But P x cos a v P 2 cos of 9 , etc., are the distances of the extremi- 
ties of the lines representing the respective forces from the 
plane of Z Y, and hence also of the equal particles placed at 
these points, denoted in the preceding sections on the centre 
of gravity by x v x„ etc. 

Hence if m represent the mass or weight of each equal 
particle, M the whole mass of them aggregated, and X the 
distance of the centre of gravity of the system from the plane 
of Z Y, we have [80] 

MX = 2 m x = m x x -f- m x^ -f- etc. 
= m \P 1 cos a x -\- P^ cos a 2 -f- etc. \— m 2 P cos a = o, 

o 



IOO ELEMENTS OF MECHANICS. 

Hence the centre of gravity lies in the plane of Z Y, perpen- 
dicular to AX, and passing through the point A. But as 
^Zwas assumed arbitrarily, this centre of gravity must lie 
in every plane which can be drawn through the point A, and 
must therefore fall at A itself, the common point of intersec- 
tion of the planes. The truth of the proposition as enunciated 
is thus apparent. 

[83.] On the other hand, if from each particle of a system 
of equal particles we draw lines to its C. of G., and suppose 
these lines to represent a system of forces applied at that 
point, these forces will be in equilibrium. For if we suppose 
the plane of Z Y in any position to pass through the C. of G. 
of the whole system, then 

2 m x 

,\ 2 m x = m 2 P cos a = o, 

and the algebraic sum of the components of the various forces 
lying along the axis of X in any position it may occupy is 
equal to zero, and hence the forces are in equilibrium. The 
axis of X may be supposed to coincide with the three posi- 
tions for the axes of X, Y, and Z successively, giving the con- 
ditions of equilibrium in [31]. 

Corollary. — Since the C. of G. of a triangle is the same 
as that of three equal particles placed at its vertices [74], it 
follows that if three forces are in equilibrium about a point, 
this point will be the centre of gravity of the triangle formed 
by joining the extremities of the lines representing the forces. 

Let the student show in a similar manner, referring to \J7\ 
that if four forces are in equilibrium about a point, that point 
will be the centre of gravity of the triangular pyramid, whose 
vertices are the extremities of the lines representing the forces. 

[84.] If any two forces applied at a point are reversed in 
their directions, their resultant will also be reversed in direc- 



STATICS. IOI 

tibn, but remain of the same value. If, moreover, the resul- 
tant of the two forces be used with a third force, applied to 
the same point, and both be reversed in direction, they will 
give a second resultant equivalent to the resultant of the three 
given forces reversed in direction ; and by continuing the 
process it may be shown that reversing the direction of*any 
number of forces applied at a point will reverse their resultant 
simply, without changing its value. 

If, therefore, a system of forces soliciting a point be in 
equilibrium, or their resultant equal to zero, they will also be 
in equilibrium when they are reversed in direction, and pro- 
duce simply a pressure upon the point. Hence, taking this in 
connection with what precedes, if we suppose all the equal particles 
of a rigid body of any form to be attracted to its centre of gravity 
by forces directly proportional to their distances from this point, 
they will be in equilibrium amongst themselves. 

PROBLEMS. 

[85.] Let the student solve the following problems illus- 
trative of section [83]. 

(a) The co-ordinates of 5 equal particles lying in the plane 
of X Fare (2, o), (o, 1), (4, 1), (5, 3), and (2,4). Required, first 
analytically, and secondly by graphical construction, the co- 
ordinates of the point, from which if lines representing forces 
be drawn to the positions of the various particles, these forces 
so represented will be in equilibrium. 

(b) Four equal particles, a, b, c, d, are taken in any posi- 
tion in one plane. Required by construction the position 
of a fifth equal particle, so that the centre of gravity of the 
whole shall be at some assumed point in the plane, arbitrarily 
taken. 

It will be observed that it is not necessary that the particles 
shall lie in one plane, as in the examples given, since section 
[83] is applicable to forces lying in any direction. They were 
so taken simply for convenience of construction. The teacher 
may give out other examples on the foregoing sections. 



102 ELEMENTS OF MECHANICS. 



Of the Mechanical Powers. 



[86.] The simplest machines that are used to render a force 
applied at one point, practically available at another, are called 
the mechanical powers. Amongst them may be reckoned the 
Lever, the Wheel and Axle, the Pulley, the Inclined Plane, the 
Wedge, and the Screw. It will be seen, however, that these 
six may be reduced in principle to two, the Lever and the In- 
clined Plane. As will become apparent in what follows, the 
mechanism of the wheel and axle merely continues the prin- 
ciples of the lever by means of ropes or cords ; the properties 
of the screw result from a combination of those of the lever 
and inclined plane ; and the wedge consists simply of two in- 
clined planes with their bases coinciding. The object of the 
following discussions will be simply to determine the condi- 
tions of equilibrium of the power or force employed, and the 
resistance to be overcome, whatever combination of mechani- 
cal powers may be used. 

When this equilibrium is destroyed, and motion ensues, 
the consideration of the effects due to this motion belongs to 
Dynamics. 

THE LEVER. 

87.] The lever is a simple rod or bar supported upon a 
fixed point, as a centre of motion, and upon which it is sup- 
posed to move freely. This point is called the fulcrum. The 
power applied, and the weight or resistance of whatever kind 
to be overcome, are supposed to be applied at other points of 
the bar. 

In a lever of the first kind, the fulcrum is between the 
power and the weight (Fig. 53). 

To determine the conditions of 
equilibrium, let the power and 
weight be supposed to act in a 
vertical direction, and to be equal 
respectively to P and W. Denote 
the lengths of the power-arm CD 




ST A TICS. 103 

and the weight-arm A C by p and w, whilst D and A are the 
points of application of the power and weight respectively. 

It is obvious that when there is equilibrium, the fulcum C 
must in its position be coincident with the point of application 
of the resultant of the two parallel forces P and W. Hence 
we must have [33] 

P:l¥::w:p, or P.p=W.w, 

as the condition of equilibrium. 

If the line E C B be drawn through C, perpendicular to the 
direction of the given forces, the similar triangles ACE and 
BCD give 

w ~ EC 
Hence P.BC=W.EC; 

or, there will be equilibrium when the moments of the power and 
iveight zvith respect to the fulcrum are equal. It will be observed 
that we have here supposed the power and weight, or resist- 
ance, to act in parallel directions. 

In a lever of the second kind (Fig. 54), the weight lies 
between the power and the fulcrum. In p 

this case also it is evident that there 
will be equilibrium only when the point ff 
C is coincident with the point of applica- 



tion of the resultant of P and IV, or when a W 

the moments of these forces with regard 
to the point C are equal. Hence we f p 

have still [34] 



P:W::w:p, or P.p=W. w, 



A 



Fig. 55- 



and it is evident the same equation will 
likewise obtain in the case of a lever of the third kind, as seen 
in Fig. 55, where the power is applied at a point between 
the fulcrum and the weight. 



104 ELEMENTS OF MECHANICS. 

[88.] When a weight is sustained by a rod or bar PP X 
(Fig. 56), the pressures upon these points p B & 

are inversely as their distances from the A "T "A 

given weight. 

For if these pressures are denoted by 
P and P lf and we conceive each point of support to become 
alternately a centre of motion, we shall have, from what has 
been shown above, 

P: W:: BP X : PP X and P x : W :: B P : PP X ; 

whence P : P x :: B P x : £P, 

as enunciated. 

[89.] In what precedes we have not supposed the levers 
themselves to possess any appreciable weight. Considering 
them, however, to be of uniform weight throughout, we may 
conceive this weight to act at their centres of gravity, or the 
weight of each arm at its centre of gravity. In a lever of the 
first kind, let g denote the weight of the power-arm, and g x 
that of the weight-arm. The moments of the two arms would 
then be 

p w 

s 2 and & 7' 

which, in connection with the moments of the power and 
weight, in case of equilibrium, gives 



Ww 

p w 

P> P +£^ = W. w + g x -, or 



w p 



2' p 

Or the weight of the whole lever might have been considered 
as acting in conjunction with the power P, at a distance from 

the fulcrum equal to ( ), the distance of its centre of 

gravity from that point. Whence we have 



STA TICS. 105 

P-P + ig + S.) (~-') = W.w, 

p w p w 

W.w-g-+g~-g x ~ 2 +g x -- 

or P-— — — £-. 

P 

But the arms being of uniform weight, 

w p 

g\g,\\p\w, or g—=g 1 ~. 

The value of P therefore becomes 

p w 

W.zv-g- + g-- 

P = #— -' 

as found before. 

In a lever of the second kind, the whole weight of the lever 
(Fig. 54) may be supposed to act at a distance from the ful- 

P 
crum equal to -. Hence, denoting the whole weight by G, 

we have, in case of equilibrium, 

P.p= IV. zv + G^, or P= W- + —. 
r '2 p l 2 

In a lever of the third kind (Fig. 55), the distance of the 

centre of gravity from the fulcrum is equal to — . Hence the 
equation of equilibrium will be 

P.p=W.w+G^, and P= '<»( W +i G \ 

r ' 2 p 

[90.] In the preceding examples, the forces have been 
supposed to be applied to a straight lever. It remains to ex- 
amine the more general case, in which neither the forms of 
the lever nor the direction of the applied forces are to be 
restricted, except that they are to lie in the same plane. 



io6 



ELEMENTS OF MECHANICS. 



Let A B (Fig. 57) be a given bent lever, and let Pand P l 
be the applied forces. Con- 
ceive a plane to pass through 
the lever and lines of direction 
of the two forces. Produce 
the lines representing the 
forces in this plane, and draw 
CE and CD perpendicular to 
their lines of direction. We 




Fig. 57. 



may now suppose P and P x to be transferred to the points E 
and D, and the moments of the two forces with respect to C 
will be P. CE and P l . CD, or P.p and P 1 .p x , and in case of 
equilibrium we have 

P.p~P x .p v 

Hence, when two forces act upon the extremities of a lever 
lying in the same plane, there will be equilibrium when they 
are to each other inversely as the perpendiculars from the 
fulcrum to their lines of direction. 

[91.] Let it be required to determine by a graphical solution 
the ratio of the two forces necessary to produce equilibrium, when 
they arc applied at the extremities of a given lever, and arc not 
parallel. 

Let A B (Fig. 58) be the given lever, and Pand P 1 the ap- 
plied forces. Producing their 
lines of direction to intersect at 
E, draw E C, and likewise CE 
and C G, respectively parallel to 
EA and E B. 

Supposing P and P l to be 
transferred to E, as their point 
of application, it is obvious that 
there can be equilibrium only when their resultant lies in the 
direction E C, and would pass through C if prolonged. But 
E C is the resultant of the two forces E E and EG; and in 
order that any two forces applied at E, along E B and EA r 




Fig. 58. 



ST A TICS. 



IO/ 



may have a resultant in the direction E C, they must plainly 
have the same ratio to each other as EF and E G. Hence we 
have, in case of equilibrium, 



P_ 

1\ 



EG 

EF' 



If one of the forces be given, the remaining one becomes 
known. 

[92.] The False Balance. — Let AB (Fig. 59) be a pair of 
scales equally balanced, but having the 

arm AC somewhat greater than B C. A_ ^c Q 

They will have the appearance of true 
scales, Avhen in reality they are false ; 
for though the two arms may be in equi- 
librium, yet if a weight Wwere placed in 



iw 



w, 



Fig. 59. 



each scale, the moments about C would be 



W.AC 



and 



W.BC. 



But since A C> B C, we should have 

W.A C> W.BC, 

and, the moments being unequal, the equilibrium would not 
continue with equal weights in each scale, showing the scales 
to be false. 

To find the true weight by means of the same balances, 
let us suppose the thing to be weighed to be placed first in 
the scale A and to balance P pounds in the other, and when 
placed in the scale B to balance P l pounds placed in A. 

If now its true weight be X, we shall have, from what has 
been heretofore shown, 



P:X::AC:BC and 
Whence PP, = X\ and 



P : X :: B C : A C. 



x= Vpp x \ 

that is to say, the true weight will be equal to the square root of 
the product of the weights in the two scales. 



io8 



ELEMENTS OF MECHANICS. 



w, 



Pa 



| 93. | The Compound Lever. — When several levers are com- 
bined in their action, so that the weight-arm of the first acts 
upon the power-arm of the second, the weight-arm of the 
second upon the power-arm of the third, etc., the combination 
is called a compound lever 
(^Fig. 60). 

To determine the condi- 
tion of equilibrium, denote 
the power-arms taken in 
order bv A A« A> ct c, aiu ^ tne 
corresponding weight - arms 
by :o, :o x , ze/ a , etc. 

The effect of the first lever 
upon the power-arm of the second in the direction of 0, is 
found from the proportion 

r.p 



w 



Fig. 60. 



w\p\\P\ Q } 



or 



Q = 



w 



the effect of the second upon the power-arm of the third in 
the direction of Q v from the proportion 



A 



/, • w, : : Si : 6, or Q l = Q 

and of the third, to raise the weight IT, from the proportion 

w, :a :: L\ : W\ 
whence, in case of equilibrium, we have 

A „A-A 'V-A-A 



or 



w- a ti = q 

/\/.A-A = " 



Hence, in any case when there is equilibrium, and the 
levers are supposed to possess no weight, the power into the 
continued product of the power-arms must be equal to the 
weight into the continued product of the weight-arms. The 
student will be able to modify the formula in practice, in case 



STA TICS. 



IO9 



the weights of the levers are not inconsiderable in comparison 
with the applied forces. 



THE WHEEL AND AXLE. 

[94-.] As has already been stated, the wheel and axle 
merely continues the action of the lever. 

Let A B (Fig. 61) be a cylindric axle, having the two arms 
5 Jll and 5 N attached to it by a rigid 
connection, and extending to the axis 
of the cylinder at S. 

The power P and the weight W 
will tend to revolve it in opposite 
directions about the axis, and the 
point 5 may be considered as the 
fulcrum of the lever M N, to which 
the power and weight are applied. 

Hence when there is equilibrium, 



L_b 



gw 



\^LL 



TTT 



\t ' / j> 



Fig. 61. 



P: IV:: SJV: SM. 

If the weight-arm be supposed to diminish in length until 
it becomes equal to the radius of the axle, which denote by r, 
we have 

P: IF:: r: SM, 

and the cord to which the weight is attached, becoming tan- 
gent to the cylinder, will wind around it in its revolution, 
thus raising the weight, which continually acts with a lever- 
arm equal to r, the radius of the axle. 

If SM should become the radius of a wheel attached to 
the axle, upon the circumference of which the power is sup- 
posed to act, and we denote by R the radius of the wheel, 
equal to SM, we have 

P: IV:: r: R. 

If we suppose the power P to be applied to a rope passing 
over the circumference of the wheel, and tangent to it at M, 



no 



ELEMENTS OF MECHANICS. 



the radius MS = R would be the power-arm ; but if the 
power produce a revolution of the wheel, its consecutive radii 
will in succession become the power-arms, and those of the 
axle the weight-arms, so that an equal lever is continued 
throughout the whole revolution. 

[95.] The Differential Axle. — The effect of the wheel and 
axle for practical purposes may be very greatly increased by 
having the two portions of the axle (Fig. 62) of different 
diameters, and the rope sustaining the weight to be raised so 
attached that when it winds upon the larger portion it will be 
unwound from the smaller. This is accomplished by causing 
the rope, which passes around a movable pulley to which the 
weight is attached, to pass on different sides of the two por- 
tions of the axle. Let the power 
P, for example, applied to the 
lever Q, which may represent the 
radius of a wheel, be made to 
complete one revolution in the 
direction PN. 

In this case one portion of the 
rope sustaining the weight would 
be taken up an amount equal to Fig. 62. 

the circumference of the larger portion of the axle, and the 
other let off by an amount equal to the circumference of the 
smaller portion. 

The distance through which the weight would be raised 
would plainly be equal to the difference of these two quan- 
tities, or, calling the radius of the larger axle R and that of 
the smaller r, every revolution would raise the weight through 
a distance equal to 

D = 27tR — 2nr — 27t{R — r). 

Observing that the rope, passing over the smaller portion 
of the axle, tends to turn it in the same direction as the power 
P by a force equal to the tension upon the rope, equal to 
\ W, acting with a lever-arm equal to r, the radius of the 




ST A TICS. 



Ill 



smaller portion, the equation expressing the condition of 
equilibrium becomes 



P.Q 



W 



W 



.*, 



or 



P = 



W(R - r) 



2 2 2 

For a given weight, therefore, the power decreases as the 
difference between the radii of the two portions of the axle ; 

and finally, if R = r, we have 

P=o, 

as should be the case, since the two portions of the rope, 
being applied to equal cylinders, would be in equilibrium of 
themselves. 

In this case, however, 

D = 2 n (R — r ) = O, 

and the weight would not move. 

It is evident that when R is but slightly greater than r, a 
very small force P might be made to raise a very great weight ; 
the motion of the weight decreasing, however, with the de- 
crease of the power, due to the difference between R and r 
becoming less. 

Example. — Let P= 10 lbs., R = io inches, r — g inches, and 
Q = 20 inches, to find the weight W that would be equili- 
brated, and the distance through which it would be raised by 
one revolution of the axle. 



the cogs of 



[96.] If there be a system of 
one work into those 
of another, they will plainly act 
upon the principle of the com- 
pound lever [93], where the sev- 
eral power-arms are represented 
by the radii of the larger wheels, 
and the weight-arms by the radii 
of the smaller, the fixed centres 
of the wheels acting as so many fulcrums 



wheels (Fig. 63) in which 




Fig, 63. 
Hence if 



112 ELEMENTS OF MECHANICS. 

etc., represent the radii of the larger wheels, and r, r v r„ etc., 
those of the smaller, the equation of equilibrium becomes 

P.R.R 1 .R, = W.r.r 1 .r 2 . 

To find the distance through which the Aveight would move, 
due to one revolution of the first wheel, it will be observed 
that the second wheel would perform a portion of a revolu- 
tion, equal to the circumference of the wheel acting upon it 
divided by its own circumference, or equal to 

2 nr r 

'tTu'r, = r; 

Since, also, one revolution of the second wheel would cause 
the third to revolve through a space equal to 

2 7t r r. 



2 7r# R, 



T 

it follows that a part of a revolution denoted by -^ would 
cause it to revolve through a portion of a revolution equal to 

r r, r.r. 

X 



R/\R, R.R,' 

and since one complete revolution of the axis to which the 
weight is attached is equal to 2 7tr 2 , the distance through 
which the weight would be raised by the portion of a revolu- 

T V 

ion denoted by -^— ~ would be equal to 

and generally, where there are any number n of wheels, we 
have 

- R, R, . . . R n _, 




STATICS. 113 



THE PULLEY. 

[97.] A pulley is a small grooved wheel made to revolve 
upon an axis, which is either fixed or movable. Hence there 
are fixed and movable pulleys. 

Let A (Fig. 64) be the fixed axis of a pulley, over which a 
cord passes, to which are applied the forces y 
P and P x acting tangentially to the circum- 
ference at the points B and C Producing 
the lines of direction of the forces to meet in 
V, it is obvious that when there is equili- 
brium the resultant of P and P lf transferred 
to V, must pass through A, the point of sup- 
port. Hence the angle Fwill be bisected, 
and consequently P must be equal to P x . 

Let Vm — Vn — P x = P represent the FlG - 64 ' 

force applied at V, and draw m s and n s, respectively parallel 
to VB and V C, and the chord BC 

Similar triangles Vn s and ABC give 

Vn : Vs\: BA : BC; 

or, since Vn represents the force P; and V s the resultant R of 
the forces P and P v which produces the pressure upon the 
axis of the pulley at A, we have, denoting the radius of the 
pulley by r, 

P: R::r: BC 

We conclude, hence, that each of the forces applied to the cord, 
is to the pressure upon the axis of the pulley, as the radius of the 
pulley is to the chord of the arc in contact with the rope. 

(a) When the forces P and P x act in parallel directions, the 
chord becomes equal to the diameter, or 2 r, and 

P : R : : r : 2 r, or R = 2P. 

(b) When the arc B C — 6o°, the chord will be equal to r, 
and 

P=R; 



ii 4 



ELEMENTS OF MECHANICS. 



and if the arc is less than 6o°, the applied force must be greater 
than R, or the pressure upon the axis, and we have 

P>R. 

(c) If Pj represent a weight Wto be raised vertically, we 

have 

P=W. 



Hence there is no mechanical power gained in the fixed pul- 
ley, and the only advantage derived from its use is in the 
convenience it affords of applying the power in a direction 
different from that in which it is designed to raise the weight, 
or move a resistance. 

[98.] The Movable Pulley.— -Let A,B, and C (Fig. 65) be a 
system of three movable pulleys, and let 
j^be the weight to be raised, or resistance 
to be overcome. The weight W being 
sustained by the cords x and y equally, the 
tension of x, or the pressure upon the pul- 
ley B, is equal to \W. This strain is again 
divided between the cords x l and y v and 
the tension of x xi or pressure upon C, is 

equal to ■§- of \ W — —^ W. In like manner 

this pressure will be divided between the 
cords x^ and y„ and we have the strain upon 
x\ equal to the applied power, equal 

P = ioi 1 - 2 W= 1 - s W; 

and it is evident that if the number of movable pulleys were 
n, a like course of reasoning would give as the equation of 
equilibrium between the power and weight, when the weight 
of the pulleys is neglected, 




P = 



1 W 

— W—-- 

2 n 2 n ' 



or 



W=2 n .P. 



ST A TICS. 1 1 5 

When ;/ becomes large, by passing to logarithms, to avoid 
the inconvenience of raising 2 to a high power, we have 

" log IV = n log 2 -f- log P f or log P = log IV— 11 log 2. 

Hence if any two of the unknown quantities P, IV, and n be 
given, the remaining one can be determined. 

If it be desired to consider the weights of the pulleys, let 
n denote the number of movable pulleys, and w the weight of 
each one of them. From the formula above, we shall then 
have for the power necessary to sustain the weight of each, 
taken in order, from that one to which the power is applied, 

www w 

A = --, A = —*, A ■= -i, A = — ; 

and to sustain the weight of the whole number, 

A+A+A+ • • • • A=^(j + j. + j-+ • • • • ^)- 
The sum of the series included in parenthesis is equal to 

( J - ?)• 

Hence the whole power necessary to sustain both the weight 
of the system and that attached, becomes 

■ u> W W — w 



W — — - H ~ = IV 



2 n 1 2 n ' 2 n 

Example. — A weight of 1000 lbs. is sustained by 8 movable 
pulleys; to find the power necessary to sustain the weight of 
the whole, when each pulley is supposed to weigh 5 lbs. 

A little consideration will show that if the last pulley A 
were raised through a vertical distance, say of one foot, it 
would free this amount of rope or cord on each of the two 
branches of the rope x and y, and the amount of rope freed 
would be two feet, or double the amount the weight would 
be raised. Hence any upward motion of A, or the weight W, 



116 



ELEMENTS OF MECHANICS. 



will give a double vertical motion to B, and in like manner 
the motion upwards of C, would be double that of B, and this 
ratio of increase would be preserved for any number of pulleys. 
Hence, if there be n pulleys, and the weight, or the pulley to 
which it is attached, be supposed to move vertically a distance 
equal to unity, we shall have for the motion of the power, de- 
noted by D, 

Hence, generally, if d denote the distance moved through by 
the weight, we shall have 

D = 2 n d. 



But from what precedes, 



Hence 



W=2 n P. 

P~~d ; 



or in case of motion, the motion of the power will be to the motion 
of the weight, as the weight to the power. 



[99.] If the cords sustaining the system and the weight to 
be raised are supposed not to be parallel, the ratio of the ten- 
sion of any portion of the cord t 
to the resistance offered by the \ z \ Si 
axis of the pulley to which that 
portion applies, will be deter- 
mined by the principle estab- 
lished in section [97], which 



plainly applicable also to a mov- 
able pulley. 

Let A j B, and C (Fig. 66) rep- 
resent such a system of pulleys, 
with radii equal to r, r v and r 2 
respectively, and the chords 
joining the points of tangency equal to c 




Fig. 66. 



and <r 2 , the points 



ST A TICS. 



f, f, and f being fixed. We then have, irom the section re- 
ferred to, 



/ : W 
A: P 



whence, by multiplying the like terms of the proportions and 
cancelling like factors in antecedent and consequent, 

P : W : : r .r 1 r 2 : cc l c l ; 

that is to say, the power is to the weight as the continued product 
of the radii is to the product of the chords subtending the arcs of 
contact. 

[100.] The following system, or combination, is more con- 
venient than the preceding. It consists (Fig. 67) of several 
pulleys, fixed and movable, requiring but a single cord. Let 
us suppose that B, D, and F are susceptible of a 
rotary motion only, and A, C, and E to be mov- 
able pulleys, whose centres are, however, rigidly 
connected, as represented in the diagram. 

For the moment, let all of the pulleys be con- 
ceived to be of the same size, and let A be sup- 
posed to ascend until it has made one complete 
revolution. There will then be, as will appear 
from section [98], sufficient rope freed from A to 
cause B to revolve twice, and this same amount 
of rope passing also over the remaining pulleys, 
C, D, F, and F, will cause each of them in like 
manner to perform two revolutions. But as A 
ascends sufficiently to perform one revolution, the 
two pulleys C and E rigidly connected with A also ^ . 
ascend sufficiently to perform one revolution on 
this account alone. Hence the revolutions of C will be equal 
to 2 -\- 1 = 3. Again, since the rope freed from the ascent of 
A gives D two revolutions, so likewise that freed from the 
ascent of C will give it two. Hence the revolutions of D will 
be 2 + 2=4. Similarly F owes one revolution to its own 



u8 



ELEMENTS OF MECHANICS. 



ascent, two to the rope freed IxomA, and two to that freed 
from C. Hence the revolutions of E will be i —J- 2 — |— 2 = 5. 
And finally, the rope freed from A, C, and E gives in each case 
two revolutions to F, which therefore performs 2 -f- 2 + 2 = & 
revolutions. To resume : 

Whilst A performs 1 revolution, 



B 


a 


2 


revolutions 


C 


it 


•2 




n 


D 


a 


4 


a 


E 


it 


5 


a 


F 


a 


6 


u 



Thus it is seen that a system of fixed and movable pulleys 
of equal diameters will revolve in unequal times ; but should the 
diameters — or, which would be the same, the circumferences — 
of the various pulleys be made in the same ratio to each other 
as the number of revolutions they make in a given time when 
supposed equal, they would then all revolve in the same time. 
That is to say, if the diameters of A, B, C, D, E, and F be re- 
spectively equal to 1, 2, 3, 4, 5, 6, the amount of rope passing 
over them, being in the saute proportion as their diameters or cir- 
cumferences, will cause them all to revolve in the same time. 

[101.] White's Pulley. — In the combination 
given above, each pulley is supposed to move 
upon an independent axis. A more perfect ar- 
rangement, however, may be found in White's Pul- 
ley (Fig. 68), which consists of several grooves, 
representing the pulleys of the preceding case, 
cut upon a single block, and working upon the 
same axis. The grooves upon the fixed block A 
are so cut that their diameters may be in the ratio 
of the numbers 2, 4, 6, etc., and those upon the 
movable block B in the ratio of the odd numbers 
1, 3, 5, etc. 

If these grooves, so related to each other as 
to their diameters, were independent pulleys, 
they would, as shown above, all revolve in the 




STATICS. II9 

same time ; and hence, though they are simply the grooves of 
the same block, since the rope passing over each tends to re- 
volve the different portions of the block with the same angular 
velocity, no resistance is offered to its moving upon a single 
axis. 

To obtain the equation of equilibrium between the power 
and weight, or resistance, in either of the cases just presented, 
it will be observed that, from the freedom of motion in all 
parts of the system, the weight is equally divided amongst all 
the ropes which sustain it; but as the power P has only to 
equilibrate the rope passing over the last pulley or groove, it 
will be equal to the tension of a single portion of the rope. 

Hence if the weight W be divided amongst n ropes, the 
power would be equal to 

P=-W 

n 



THE INCLINED PLANE. 

[102.] The mechanical advantage to be derived from the 
use of the inclined plane is easily understood, from the for- 
mulae already found, in section [21], Ex. 4, of the application 
of the principle of the parallelogram of forces ; in which it 
was shown that when the power is applied parallel to the 
length of the plane, and there is equilibrium, it is equal to the 
weight multiplied by the height of the plane, and divided by 
its length ; and when applied parallel to the base, is equal to 
the weight multiplied by the height of the plane, divided by 
the base. 

In either case, therefore, since the power is less than the 
weight, except in the latter case when the height of the plane 
exceeds the base, there is a gain of mechanical effect which 
may be estimated by the formulas referred to. It should be 
observed, however, that these formulas were derived upon the 
hypothesis that there was no friction, the consideration of 
which will follow hereafter. 



120 ELEMENTS OF MECHANICS. 



THE WEDGE. 



[(03.] The wedge is commonly used to separate the parts 
of a body by means of a force applied to its back, tending to 
urge it forward between the particles or fibres of the body, 
which generally separate in advance of the edge of the wedge. 
The wedge is, however, used for other purposes ; as, for ex- 
ample, to tighten the tenon which fits into a mortise, to raise 
great weights through small distances, etc. To determine 
the pressure produced by the sides of the wedge in a direc- 
tion perpendicular to the side or face, let A F (Fig. 69) repre- 
sent the impulsive force or pressure upon 
the back of the wedge. Resolving this B ^///.a --/^c/ j 
force into its two components A P and 
A P lt these components will represent the 
effective force of the wedge upon a resist- 
ance acting perpendicularly to its sides or 
faces. Comparing the similar triangles 
A PF and B D C, whose sides are mutually 
perpendicular, we have Fig. 69. 

B C: CD :: A F : AP; 

and multiplying antecedent and consequent by C E, we get 




BC.CE: CD.CE :: AF: A P; 

that is to say, the applied force is to the effective force of the 
wedge, in a direction perpendicular to its faces, as the surface of 
the back of the wedge is to the surface of one of its faces. 

This of course is a theoretical result, in which no regard 
has been paid to friction, which is, however, necessary to the 
utility of the wedge, in preventing it from flying out on ac- 
count of the pressure upon its sides. 

THE SCREW. 

[104.] Referring to section [21], Ex. 4, of the case of a body 
upon an inclined plane, we find that the equation of equili- 



STATICS. 



121 



brium between the power and weight, when the former is 

Wh 

applied parallel to the base of the plane, is/= — j- . 

Before attempting to apply the formula to the case of the 
screw, it will be well first to consider what the effect would 
be if, instead of applying a force to the body parallel to the 
base of the plane, we suppose the plane, moving upon rollers 
(Fig. 70), to be urged against the body, which, as shown in the 
diagram, may be supposed to be capable of moving vertically, 
without friction, upon the rollers placed against an immova- 
ble obstacle 6". 

Let be the centre of gravity of the body, and let o c rep- 
resent its weight. Draw o a per- 
pendicular to the plane, and o b 
parallel to a c, parallel to A B, the 
base of the plane, and complete 
the parallelogram ab c. 

The pressure of the body 
upon the plane when there is 
equilibrum, produced by a force /applied against B C parallel 
to the base of the plane, will be on, and the reaction of the 
plane against the body will be an equal force a in the oppo- 
site direction. It is obvious that in case of equilibrium the 
force /"balances the force ca, the component of the pressure 
oa parallel to the base of the plane. But if we suppose the 
plane to be urged by a force / equal to m n or greater than 
a c, the reactionary force of the plane through a would sustain 
a weight represented by n. Hence any force greater than 
a c would give a vertical motion to W. 

From the triangles ABC and a c we 
equilibrium, 




Fig. 70. 



have, 



in case 



of 



ac : oc :: B C : AB, 

and f — 



or 
Wh 



f: Ww h : b, 



as in the case of a body impelled against the plane by a force 
parallel to the base. 




122 ELEMENTS OE MECHANICS. 

The application of the above to the case of the thread of a 
screw urged against the groove in the nut, or the groove 
formed by the progress of the screw through wood, will be 
easily understood from the following illustration. Let a piece 
of paper, cut in the shape of a right-angled triangle, be sup- 
posed to be wrapped upon a cylinder BD (Fig. 71), so that 
the hypothenuse of the triangle would describe 
the thread of a screw 7/1/1, op, etc., around the 
cylinder. If the paper be then unrolled one 
complete revolution, forming the inclined 
plane A C, and the cylinder be turned in the 
direction of the arrow PS, the plane A C, 
or thread of the screw, would be urged 
against the body W in the direction of the " FlG 
base of the plane A B. 

If, instead of the body IV, the thread of the screw should 
come in contact with the grooves of the nut, the only difference 
in the nature of the resistance offered would be that the nut 
would offer a resistance to vertical motion throughout the 
whole extent of the thread it covers. 

It is plain, therefore, that if we consider the nut to move 
without friction, and to be pressed down by a given weight 
or resistance to be overcome, the formula above would con- 
tinue to apply ; and if the power urging the thread at the point 
C on its circumference be denoted by P, we have 

Wh _ IV. B C 
b ~ AB ' 

or calling the weight or resistance to be overcome R, the dis- 
tance between the threads of the screw equal to B C = d, and 
the circumference of the cylinder upon which the threads are 
cut equal to A B = c, this equation becomes 

P=R-, or P = P^ 

c a 

Should the power be applied to a lever OP, equal to 
11 times the radius of the cylinder, the force exerted at the 



STATICS. 123, 

point C would be increased n times, and the equation would 
become 



But since the circumferences of circles are as their radii, the 
circumference described by the power P would be equal to 
;/ times the circumference of the cylinder, or equal to nc. 
Hence the last equation may be written 

_ P.cir.ofP 
R = d ' 

It should be observed that this formula is independent of 
the circumference of the screw cylinder, the gain in the rela- 
tive leverage of the power in a smaller cylinder being counter- 
acted by the increased steepness of the thread, the distance 
between the threads remaining the same. 

The formulas above, derived upon the hypothesis that 
there is no friction, are of course only theoretically true, 
affording a greater or less degree of approximation to practi- 
cal results as the friction is less or greater, in comparison with 
the resistance to be overcome. This approximation will also 
be greater the less the inclination of the threads of the screw,, 
as will be manifest by considering the analogous case of the 
inclined plane ; for the greater the inclination of the plane the 
greater the pressure of the body when held in equilibrium by 
a force parallel to the base, and consequently the greater the 
friction to be overcome. 

(a) Required the weight that can be raised by a screw, 
when the distance between the threads is equal to |- of an inch, 
the length of the power-arm 2 feet, and the power P = 10 lbs. 

(b) The resistance to be overcome being 2842 lbs., the dis- 
tance between the threads -J- of an inch, and the power 12 lbs.,, 
to find the power-arm of the lever to be employed. 



124 



ELEMENTS OF MECHANICS. 




Fig. 72. 



[105.] Friction. — If we suppose a body B (Fig. 72) to be 
placed upon an inclined plane A C, it will be necessary to ele- 
vate the plane to a greater or less height above the horizontal 
base A D before the body will just be on the verge of mov- 
ing down the plane. This will be the case however polished 
the surfaces may be. If they were perfectly smooth and sup- 
posed to be entirely without friction, the motion of the body 
would take place for the very slightest elevation of the plane. 
Owing to the friction of the surfaces 
in contact this result is, however, 
never attained in practice. Resolv- 
ing the weight of the body B W—W 
into its components BP — P and 
Bf—f, respectively perpendicular 
and parallel to the plane, the force 
tending to produce motion down the plane is equal to Bf=f. 
In case of equilibrium, when the plane is at the greatest 
elevation at which no motion will ensue, the friction will 
equilibrate the force /, and is therefore equal to it. 

But, from similar triangles, 

W: P:: A C: A D : 1 : cos a, or P—Wcosa, 
and 
W : f : : A C : C D : 1 : sin a, or /= IV sin a — P tan a. 

If the w r eight of the body be increased, and consequently its 
perpendicular pressure upon the plane equal to P— Wcos a, 
experiment shows that the angle of elevation at which the 
body will begin to move down the plane will remain un- 
changed. 

But for any given angle at which motion will ensue, the 
force down the plane is directly proportional to the pressure ; 
hence also the friction which at the given angle, a say, equili- 
brates this force is directly proportional to the pressure. 

If we denote the ratio of P to /, or of the normal pressure 
to the friction, by F, we have 



F = ^ = tan a 



and f— P tan a — F.P. 



STATICS. 125 

The angle a is the angle of friction, and, as stated above, is 
constant for the same body, and independent of its weight ; 
and F, being the coefficient of the power necessary to produce 
/, is called the coefficient of friction. 

If F, be supposed to denote the friction due to the unit of 
normal pressure, /= P. F l will represent the friction due to 
the pressure P, the friction being, as already stated, propor- 
tional to the pressure. 

Experiment also goes to show : 

1. That with a given normal pressure the friction is in- 
dependent of the surface in contact. 

2. That it is also independent of the velocity, if the 
body be in motion. 

[106.] Rolling Friction. — The friction resulting from the 
rolling or revolution of one body resting upon another, as, for 
example, when a cylinder rolls upon a plane, or a cylindric axis 
revolves within a hollow socket, is called rolling friction. 
In this case there is simply a line of contact between the sur- 
faces, and the amount of friction is much less for the same 
pressure than in the case of sliding friction, and the above 
laws are not applicable. 

[107.] The values of F, the coefficient of friction, for various 
substances have been determined by experiment, and the fol- 
lowing table, taken from Parkinson's " Mechanics," gives ap- 
proximate values in several cases of friction at starting : 



Wood upon wood (without oil), 


F= 


•5 


" (with oil), 


F = 


• 2 


Wood upon metal (without oil), 


F= 


• 6 


" (with oil), 


F= 


• 12 


Metal upon metal (without oil), 


F = 


•18 


" (with oil), 


F = 


•12 



126 ELEMENTS OF MECHANICS. 

When a cylinder of wood rolls upon wood, with a single 
line of contact, the value of F as stated by the same authority 
is ^ 

From the discussion of section [105] it is seen that 
F= tan a; hence to find practically the coefficient of friction 
between any substances, it will only be necessary to place a 
body of the one substance upon an inclined plane of the other, 
and note the value of the inclination a when motion just be- 
gins. The tangent of this angle gives the value of F. 

It has been estimated that the friction of the wheels of a 
car upon the rails is about y-gVo" P ar ^ °^ * ne we ight of the 
train, and that upon the axles about the same, and conse- 
quently the whole resistance from friction to be overcome 
about -g-J^ part of the load. 

It should also be observed that the measure of the friction 
of a body just bordering on motion, or statical friction , differs 
in value from that when the body is actually in motion, or 
dynamical friction ; and though the coefficient of friction in 
each case is constant, the dynamical is less than the statical. 



Principle of Virtual Velocities. 

[108.] If we conceive a body, or system of bodies, rig- 
idly connected, and held in equilibrium by any number of. 
applied forces, to be moved in any manner an indefinitely 
small amount by some extraneous force, the points of applica- 
tion of the forces will also move through an indefinitely small 
distance, dependent upon their positions and the amount and 
character of the general displacement. The distance through 
which the point of application of any force moves is called 
the virttial velocity of the point, and the projection of this dis- 
tance upon the line of direction of the force, in its original 
position, is the virtual velocity of the force. This will be made 
plainer by a diagram. 

Let a (Fig. 73) be the point of application of one of the 
forces P before any disturbance of the equilibrium, and let a x 




STATICS. 12? 

be the position of the point after a very slight displacement. 
Then a a x will be the virtual velocity of a, and a m, its projection 
on Pa, will be the virtual velocity of the force P. 

If P x be a second force whose point of application a 2 , by 
the same disturbance, is moved to 
a z , then a 2 n will be the virtual ve- 
locity of P x . It will be observed, 
therefore, that the virtual velocity 
may lie either in the direction of 
the action of the force or in the 
opposite direction. In the first FlG - 37 * 

case it is regarded as positive, in the latter as negative. 

The product of any force by its virtual velocity, as, for ex- 
ample, P. a m, is called the virtual moment of the force, though 
it will be perceived that the term moment has here no relation 
to its significance as heretofore used. 

When any machine or system of forces in equilibrium is 
very slightly disturbed, without breaking the rigid connec- 
tion of its parts, it can be shown that the following general 
proposition in relation to the virtual moments of the forces 
will be true, viz. : 

The algebraic sum of the virtual moments of a system of forces 
in equilibrium will be equal to zero. 

This is what is termed the principle of virtual velocities, or 
equation of virtual velocities, the illustration of which in a 
few cases will be given, though it is so general in its character 
as to be very fertile in its results, and affords a basis for a work 
on mechanics, as in the "Analytical Mechanics" of Lagrange. 

The general proof of this principle, besides being some- 
what difficult, is in the present work unnecessary. Its proof, 
however, for a system of forces in one plane will be here 



[109.] If we suppose a plane in equilibrium, under the action 
of several forces lying within it, to be displaced an indefinitely 
small amount, the motion will either be rectilinear or rotary, 
or a motion both of translation and rotation. It will be shown 



128 



ELEMENTS OF MECHANICS. 




that when the displacement is very slight either of the three 
may be the result of a rotary motion simply. 

Let MN(Fig. 74) be the plane, and Pand P x two of the given 
forces applied at a and a„ and sup- 
pose that with a small displacement 
by a simple rotation, or by a combined 
motion of rotation and translation, the 
point a moves to a lt whilst the point 
# 2 moves to a z . From a and a 2 erect 
perpendiculars to a a t and a 2 a 3 , meet- 
ing in the point O. It is obvious that 
when a a 1 and a 2 a 3 are indefinitely FlG - 74- 

small, the motion of the points a and a 2 would have resulted 
from a revolution about O as a centre, called the instantaneous 
centre of rotation. 

Again, if the distance to the centre O becomes indefinitely 
great, the lines a a t and a 2 a 3 would become parallel and the 
motion rectilinear simply. 

Now let P(Fig. 75) be a force applied at B, which, by a 
small displacement, is supposed to reach m by a revolution 
about O as an instantaneous centre of rotation. Draw O A — p 
perpendicular to PB produced, O B = R, and Om prolonged 
to n ; also B E — d perpendicular to O 7n, and E C perpendicu- 
lar to B n. If we now suppose 
the displacement or angle B Om 
to be indefinitely small, the angle f $ 
EnC will become equal to OB A, 
and the two triangles E C n and 
ABO similar. But at this limit 
also BE — d, the sine of the 
angle B O m, becomes equal to 
the arc B m, the velocity of the 
point of application, and B C = v to the virtual velocity of P. 
Since, also, the triangle E B C is similar to the triangle E C n, 
it is similar to A B O. We have, therefore, the proportion 




BE : BC '■:■: OB : OA, 





STA TICS. 


129 


or d : v : : 


R : p, and == d. 

P 




Similarly any other force P x would give 




or, by division, 


vp x R d 
v x pR x ~ d; 




But it is evident that the small arcs d and d y are 


to each other 


as the radii R and R x 


, or 

R d 

K " < 




Hence vp x ■■ 


= v,p, and - = --, 
/ A 




and, by analogy, for 


any number of forces 




V 


V l V l V z 





/ A A A 
a constant ratio. 

But since by hypothesis there is equilibrium in the system 
of forces, we have from section [61], the point being consid- 
ered as a centre of moments, 

2Pp = Pp +P zPl + P %P% + P 3 p 3 + etc. = a 

Multiplying both members by m, 

m*2 Pp — P . nip + P x . mp, -f- P 2 . mp 2 + P 3 . mp % + etc. = o, 

or />*/ + P 2 ^ + P 2 ^ 2 + P 3 v 5 + etc. = 0, ^P^ = o ; 

which conforms to the enunciation of the proposition as applied to 
forces lying in one plane. 

The student should carefully observe here that if the prin- 
ciple of virtual moments, or the last equation, be assumed as 
true, we could, by reversing the process and substituting for 
7J, v v etc., their values, derive the equation 

2Pp = o. 



I30 ELEMENTS OF MECHANICS. 

Hence when the principle of virtual velocities obtains, it be- 
comes a condition of equilibrium of the system. 

If, therefore, there be two forces in equilibrium, as the 
power and weight, or resistance, in the examples under the 
mechanical powers, a condition of that equilibrium would 
also be 

2Pv = o; 

from which condition, as we shall see, the relation between the 
power and weight may be determined. 

EXAMPLES. 

{a) When P and W balance each other, in the case of the bent 
lever A C (Fig. 76). 

Since, in case of equilibrium, we have 2 Pv — o and the 
virtual moments of Pand Whave 
opposite signs, there results, pre- 
serving the same notation as in 
the preceding article, 

P v 

Pv= Wv v or -777 =-• 
1 W v 

But from the same article, Fig. 76. 



v p 

Hence when the perpendiculars upon the lines of direction of 

v 
the power and weight are known, the ratio - is known, which 

is also the ratio between the power and weight. 

P 
The ratio -- will depend for its value upon the values of 

R and R v and the angles the directions of the forces make 
with them. Thus if R and R x be supposed to be given, 
and also the angles PA O = a and W C O — /?, we should 
have 

p = R sin a and p x = R x sin /?, 




ST A TICS. 



131 



and 



P 



P_ _ R x sin p 

W ~ R sin a ' 



a known ratio, 

(£) 7#<? £tf.S7? 0/ £#<? single movable pulley when the cords are 
not parallel (Fig. 77). 

Let the centre of the pulley 0, to which the weight is at- 
tached, be raised to the position O,. 
The virtual velocity of the weight 
will be X ; and the virtual veloci- 
ty of the power, the amount of rope 
or cord freed from the two branch- 
es A B and A C. Let m and v be 
the points of tangency, where the 
cord quits the pulley, in its two 
different positions. Draw O m, O x v, 
and O x n parallel to O m and cutting 
the cord at w. Prolong B in to z. 

The angle iv O l v will be equal 
to w B n, and when the displace- 
ment is very small the tangent viv is equal to the arc vx, 
and pvw =pvx = A m. In that case, also, B w = B n. Hence 
the cord Bp will be less than the cord B A by the portion mn. 
Therefore m n is the amount of cord freed for the power by 
the branch BA ; and as an equal amount is due to the branch 
A C, the virtual velocity of P will be 2 m n. 

Hence the equation 2 Pv = o becomes 

P 0, 
W ~ 2.mn 




Pv = W\ 



or P, 2 m n = W.O O v or jjr = 



But since O m and O l n are parallel, 

00, zO 

~~ zm 



mn 



and since the triangles Omz, Oms are similar, 

zO Om 



z m 



sm 



132 


ELEMENTS OF MECHANICS. 


Hence 


P 00, zO Om Om 

W~ 2 .mn ~ 2 . z m ~ 2 . s m ~ my 



On 


= 0, cos 


/?, and 


•*• 


PX 0n = 


W. X m, 




Pcos/3 = 


Wsvsx a, 




P 

W~ 


sin a 
' cos ft 




and this corresponds to the relation between the power and weight 
already deduced in section [99]. 

(c) The case of motion upon an inclined plane (Fig. 78). 

Supposing the diagram to be drawn as given, let the weight 
move from O to V The virtual velocity of the power will 
be n, and of the weight X m. But 

X m — X sin a \ 



or 



and 

Fig 78. 

If the power be applied parallel to the length of the plane, 
= o, cos § = I, 
P sin a B C h „ Wh 

and w = -T~ = -AC = -r or p =— > 

as before determined, section [21]. 
If P be applied parallel to the base, 

p = * f 

. P sin a BC h rTT h 

and -jjz = = -j- 3 = T , or P= W T • 

W cos a AB b b 

The above examples are sufficient to illustrate the prin- 
ciple of virtual moments, but the student would do well to 
make the application also to other cases of the mechanical 
powers presented in former sections. 



PART SECOND, 
DYNAMICS. 



Of Motion in General. 

[110.] By the motion of a body is understood the act of 
changing its position with reference to some fixed point con- 
ceived to exist in space. Motion is the result of overcoming 
the inertia of matter by the application of force ; a change of 
place is the result of motion. Motion is either absolute or re- 
lative : absolute when it is referred to some fixed point of 
space, and relative when it refers to a changing of position 
with regard to some other body in motion. 

It is possible to conceive of relative motion without ab- 
solute or real motion. Thus if a body near the earth's sur- 
face be supposed to remain stationary in space, its relative 
position to bodies on the earth's surface would be continually 
changed by the motion of the earth. 

The rate of motion, measured by the space passed over by 
the moving body in an assumed unit of time, is termed velocity. 

Three general laws of motion obtain, which may be stated 
as follows : 

(a) Every body will continue in its state of rest or motion, 
as the case may be, unless compelled to change that state by 
the action of some extraneous force. This obviously results 
from the inertia of matter. 

(b) The change of velocity or motion is proportional to 
the effective impressed force. This will appear from section 
[4, Cor. 3]. 

(c) Action and Reaction are always equal and contrary to 
each other.* 



134 ELEMENTS OF MECHANICS. 

It is plain that no force can exert itself against that which 
can offer no resistance. There could, for example, be no ac- 
tion of an impressed force against a vacuum. The resistance 
is simply the measure of the force exerted, for the resisting 
body can destroy the action of an applied force only so long 
and to the extent that it can offer a resistance. As soon as 
the action of the impressed force gives a greater velocity to 
the body receiving it than that of the force itself, it ceases to 
act, and there can be no longer either action or reaction. 
But up to that time the applied force could act only so far as 
there was a resistance to be acted upon, equal in amount to 
itself. In other words, the resistance offered, or reaction, is 
developed by the action, and receives it. 



Of Rectilinear Motion, 



[ML] Uniform Motion. — When a material point, in mov- 
ing, passes over equal spaces in successive equal times, the 
motion is said to be uniform. If, in addition to this, the motion 
takes place in a straight line, the result is a uniform rectilinear 
motion. It is evident that the space accomplished by the mov- 
ing body with a given velocity V, as a uniform velocity, will be 
directly as the time of its motion. Hence if V denote the 
velocity, that is to say, the space passed over in the assumed 
unit of time, 2 V will be the space for 2 units of time, 3 V for 
3 units of time and generally t V for t units of time. Or de- 
noting the space for the time t by 5, we have 

5= Vt. 

s s 

Whence, also, V = - and t = -=- 

[112.] Uniformly Varied Motion. — A constant force is 
one which, during its action, remains unchanged in its intens- 
ity. If we suppose a body to be moved by the action of such a 
force, its inertia being the same at every instant, whatever its 



DYNAMICS. 135 

velocity [2], the increase of velocity in every indefinitely small 
but equal time will be the same. The motion will therefore 
be a uniformly accelerated motion. Should the body be already 
in motion, and the constant force tend to retard its motion, 
the inertia of the mass still remaining constant, its retardation 
would be uniform, and the motion a uniformly retarded motion. 
Since, moreover, any force is capable of destroying in an in- 
stant of time as much velocity as it could impart to the same 
mass in the same time, it follows that the retardation will be 
in the same degree as the acceleration. 

Hence if a body be supposed to move over a given space 
in a given time t, under the influence of a constant force, and 
to acquire the velocity V, if the motion be supposed to be 
reversed in direction, with an initial velocity equal V, the time 
required for the same constant force to destroy this velocity 
would be equal to the time which was required to impart it. 
Likewise, the velocity being destroyed by the same gradations 
by which it was accumulated, the average velocity in either 
case, equal to half the sum of the extreme velocities, equal to 

, will be equal, and hence the spaces accomplished, equal 

to that which would be due to the average velocity as a uni- 
form velocity, would also be equal. This will be rendered 
still more obvious by the following investigation. 

[113.] We shall now proceed to derive a formula expressing 
the relation between the space and time in a uniformly varied 
motion. 

Let a material particle, solicited by a constant force, 
be supposed to move from 

.toJ (Fig. 79) in the time /. %:£:*.% **■ *? : % 
Suppose the whole time t to 
be divided into n equal inter- 
vals of time, each equal to i\ whence n i= t. 

Let the spaces from a to b, b to c, etc., be the spaces per- 
curred by the moving body within the successive intervals of 



13^ ELEMENTS OF MECHANICS. 

time denoted by i ; the body being supposed to start from a, 
with a velocity equal zero. Let g be the velocity acquired at 
the expiration of the first interval, when the particle arrives 
at b. Then, section [4, Cor. 3], the velocities at c, d, e t . . . . 
X„X will be respectively 2g, $g, 4g, . . . . (n — i)^-, ng, as 
represented in the diagram.. 

But the space a b is percurred with a velocity greater than 
the initial velocity for the space, and less than its final velocity 
at the point b. Hence if the particle be supposed to move uni- 
formly with the initial velocity at a for the interval i, it will 
accomplish a less space than a b, but if with the final velocity 
at b a greater space than a b. Similar remarks are applicable 
to the spaces b c, cd, etc. 

It follows, therefore, that the whole space a X= S, accom- 
plished in the time t, will be greater than the sum of the spaces 
passed over in the successive intervals with the initial veloci- 
ties belonging to them, and less than the sum of the spaces 
due to the final velocities. In other words, we shall have 

S>o.i + g.i+2g.i+3g.i+ ....(»- i)g.i, 

and 5 <g.i+2g.i+$g.i-\-. . . . (n — i)gi+ngi; 

or S>g.i\ 0+1+2+3+ ....(»- i)\ r=gi.\^-^\n f 

and S <gi(i+2 + i+ .... „)=^!±Ij w . 
The true value of S, therefore, lies between the expressions 
g.i[——)n and g-i\^~- )«, 



whatever may be the value of n. But when n becomes indefi- 
nitely large, these two expressions become equal in value. 
Therefore, under the hypothesis that n is infinite, the true 
value of 5 will be expressed by either. 

Hence S~<rt[- \n—gi. — ~ni. — = /. — ■• 



DYNAMICS. 137 

But gn is the final velocity at X acquired in the time t. De- 
noting this velocity by V, we have 

the space in terms of the final velocity and time. 

Again, if /denote the velocity acquired at the expiration 
of any assumed unit of time, the velocity acquired in the time t, 
section [4, Cor. 3], will be 

V = ft. 

Substituting this value of V above, we have 

S - \ft\ 

the desired formula, in which /, the velocity acquired in a unit 
of time, is called the acceleration of the moving force ; or, since 
[4] constant forces are as the accelerations they are capable 
of producing in the same mass in the same time, / is some- 
times, though improperly, called the accelerating force. 

It will be seen, therefore, that the space accomplished in a 
given time t is proportional to the square of the time, and 
also to the acceleration the force is capable of inducing in a 
unit of time, or to the accelerating force itself. 

If / = 1, equal to one unit of time, 

S = if, 

equal the space accomplished in the unit of time. 

Hence, from the formula S =$■//*, we conclude that the 
space passed over in a time t is equal to the square of the time mul- 
tiplied by the space accomplished in a unit of time. 

[114.] Again, the space passed over in n units of time being 
equal to ifn 2 , and in (// — 1) units of time equal to if(n — i) 2 , 
the space accomplished within the nth unit of time, equal to the 
difference of these two expressions, will be given by the gen- 
eral formula 

\f\n*^{n-if\^\f(?n-Y). 



138 ELEMENTS OF MECHANICS. 

Making n successively equal to 1, 2, 3, 4, etc., and denoting 
the spaces for the corresponding units of time by S lf S 2 , S 3t 
etc., we have 

S l = ifXh S."=i/X3, ^ = i/X5, St = ifX7, etc. 

Whence S 1 : S 2 : S s : S v etc. = 1:3:5:7, etc. ; 

that is, the spaces passed over within successive units of time are 
as tlie odd numbers 1, 3, 5, 7, etc. 

[H5.] Squaring the equation V — ft, and dividing the re- 
sult, member by member, by the equation 5 = \ff 

we have — = 2/ and F= V2/j, 

an expression for the velocity acquired, in terms of the space 
passed over and the acceleration for the first unit of time. 

[116.] If we now conceive the motion of the particle to 
take place in opposition to the constant force from X to a> 
in virtue of an impressed force, and with an initial velocity 
equal to V, the velocity acquired in moving from a to X, it 
will have its velocity continually diminished by equal decre- 
ments of velocity in equal times. These decrements of velocity, 
equal to the corresponding increments of velocity in the ac- 
celerated motion of the particle from a to X, would constitute 
the common difference of the same series of varying velocities 
as in [113], but reversed and forming a converging instead of 
diverging series. It is obvious, therefore, that proceeding 
according to the method of the preceding investigation, we 
should in like manner have for the space accomplished in any 
given time t, the formula 

V 
S = -f, 

the velocity V being the initial instead of final velocity, and 
the final velocity being equal to zero. That is to say, the 
space passed over by a body in a given time under the action of a 



DYNAMICS. 139 

constant force, is just equal to the space it would accomplish in the 
same time whilst moving in opposition to the same force, under an 
impulse imparting to it an initial velocity equal to that which the 
force is capable of producing in the body in the given time. 

[117.] If we denote by /the amount of velocity lost in the 
unit of time, which is equal to the acceleration the same force 
would produce in the same time, we shall have 2/ for the loss 
of velocity in 2 units of time, 3/ for the loss in 3 units of time, 
and, generally, ft for the loss of velocity in the time t. But in 
the time t the whole initial velocity V\s> lost [116]. 

Hence V=ft and S=iVt = ±ft\ 

the same formula as when the force is an accelerating instead 
of retarding force. 

This formula is therefore applicable both to uniformly ac- 
celerated and uniformly retarded motion. 

[118.] We have thus far supposed the particle to be moved, 
under the influence of a constant accelerating force, from a 
state of rest. If we conceive it to be already in motion, and 
to have a velocity u before the accelerating force begins to 
act upon it, the space accomplished in its subsequent motion 
'will be due both to the velocity u and that generated by the 
accelerating force. It is plain, however, that it is immaterial 
how the velocity u is generated. Let, then, the same accel- 
erating force under whose influence the particle is moved be 
supposed to have generated the velocity u in a time t„ whilst 
moving it through a space S t : In this case we have 

u=ft, and S^ift;. 

Let 5 be the space accomplished, after the velocity u has 
been acquired, and / the additional time of the motion. Then 
we have, for the whole space from the beginning of the motion, 

s+s^ifit + ty^iff+ftt. + ift;, 

or 5 = iff +ut; 



140 ELEMENTS OF MECHANICS. 

also, V = f(t l + 0=/'. +ft=zu+ft. 

If the force act in direct opposition to the motion already 
acquired, tending to destroy the velocity u, it is obvious these 
equations will become 



If, moreover, the time t become infinitesimal, the ratio 
ut u 



S = ut — iff and V— u — ft. 

:over, the time t become infinitesimal 

= oo , whence iff is infinitely small in comparison 

2y * 2/ t 

with u t, and may without error be neglected, and we have 

S = ut; 

that is, for any indefinitely small time the velocity in accelerated 
or retarded motion may be considered as uniform, the finite initial 
velocity being infinitely great in comparison with that accumulated 
in an infinitesimal time. 

[119.] From the equation V —u—ft, in which V is the 
terminal velocity due to the action of the force for the time /, 
with u as an initial velocity, we have 

u-V 



f 

If we suppose / to denote only a portion of the time the par- 
ticle is capable of moving against the opposing force, we 
shall have for the portion of space S v due to this time, 

~ , „ , u 2 — u V , In — V 
5 2 = ut - \ff = — - if} 



f ZJ ^ f 

or, by reducing, 

2 ~ 2/ 2/ 

If, on the other hand, / denote the whole time the particle is 
capable of moving against the constant force, until the initial 



DYNAMICS. 141 

velocity u is totally destroyed, the terminal V would become 
equal to zero, and the whole space, 5, say, would be 

2/ 

Denoting by S 3 the space remaining to be described when 
the terminal velocity is V, we have, therefore, 

V' 



^ = 5,- S 3 = — , or V=V2fS % . 

Whence it appears that the velocity of any body moving in 
opposition to a constant force ; at the expiration of a given time t y 
is proportional to the square root of the space which remains to be 
described before the motion of the body ceases. 

This result could have been rendered obvious by observ- 
ing that the velocity at any point in the path of the retarded 
motion [112] is equal to the velocity at the same point in the 
path of the accelerated motion ; but this latter velocity is 
given by the equation V= V 2fS, in which 5 is the space 
passed over, which is the same as the space which remains to 
be passed over in the retarded motion. 

[120.] The formulae deduced in the preceding articles are 
such as were derived from considering the forces applied to 
be of uniform intensity, or constant forces. They are, there- 
fore, general in their character when restricted to forces of this 
nature. The force of gravity for all ordinary distances from 
the earth's surface differs imperceptibly from a constant force, 
because the difference of these distances is so inconsiderable 
in comparison with the whole distance to the earth's centre. 
The force urging a body down an inclined plane [21] is like- 
wise of this character. 

If we assume the acceleration that the force of gravity is 
capable of giving to a mass, in one second of time, to be rep- 
resented by^*, the number of seconds of the fall by /, the space 
fallen through by S, and the final' velocity acquired at the 



142 ELEMENTS OF MECHANICS. 

expiration of the given time t by V, we have for the general 
formulae applicable to the case of falling bodies the following: 

V=gt, V=\ / ~2^ r S, and S=igt\ or t = \/il 

g 

The quantity g can of course be determined by experiment 
alone, and will vary slightly according to the latitude of the 
place, owing to the variable nature of the force of gravity for 
different latitudes. To find the value of g, let us suppose, for 
example, that in the latitude of New York careful observa- 
tions show that a body will fall through a space of 16.08 feet 
in one second. Making t= 1, and -5 = 16.08 feet, and the for- 
mula S = \gf becomes 

16.08 = \g, or g — 32.16 feet per second. 

That is, at the latitude of New York a body acquires in fall- 
ing one second a final velocity, which, as a uniform velocity, 
would in the next second carry it through a space equal to 
32. 16 feet, or 32-J- feet nearly ; which latter value may be as- 
sumed as a close approximation. 

This value of g substituted in the preceding formulas gives 



F= 3 2i/, V= V6 4 iS, 
or, as an approximation, 

or 



V--=SVS and 5 = 16.08 t\ or / = '76 < 



These formulas can now be conveniently applied to the 
solution of examples with regard to falling bodies. 

The true method of determining the value of g with accu- 
racy, will be given under the discussion of the pendulum, in a 
subsequent portion of the work. 

[121.] Recurring to section [6], we have for the measure 
of a force F the expression 

F=M--' 



DYNAMICS. I43 

and for any other force, supposed to act for the same time / 
upon the same mass M, the expression 

V 

.\F:F X :: V \ V v 

Hence any two forces may be compared with each other 
by having reference to the velocities they will impart to the 
same mass in the same time, these velocities being relative 
measures of the forces; the time t in the case of any force 
which is not constant being assumed to be indefinitely small, 
as explained in the section referred to. 

The value of g, therefore, found in [120] may, in com- 
parison with the acceleration in one second of time due to any 
other force of uniform intensity, be assumed to represent the 
force of gravity. The force which urges a body down an in- 
clined plane, being constant, will be to the force of gravity as 
the acceleration due to its action for one second is to g, the 
acceleration due to gravity for one second. 



Of the Motion of Bodies upon Inclined Planes. 

[122.] We have seen, section [21], that the weight of a body 
resting upon an inclined plane may be resolved into two compo- 
nent forces, one producing simply a pressure upon the plane, 
and the other tending to urge the body down the plane, with 
a constant intensity equal to 

WJi 

f - r ' 

In this equation/ and W both denote pressures ; but if / be 
assumed to represent the acceleration due to the force down 
the plane in one second, then may H^also be assumed [121] to 
represent the acceleration in one second due to gravity, here- 



144 ELEMENTS OF MECHANICS. 

tofore assumed equal to g, these accelerations being in the 
same proportion as the moving forces. Hence 

£- = £=* or f-^ 

w g r J ~ r 

in which equation / is the acceleration due to the force f 
down the plane, and not the force itself, but its relative meas- 
ure, the forces f and W standing related to each other in the same 
ratio as the accelerations f and g. 

[123.] The space through which a body will fall in a given 
time free from resistance, is to the space it will describe upon an 
inclined plane, in the same time, as the length of the plane I is to 
its altitude h. 

Denoting by 5 and 5„ respectively, the spaces accomplished 
in a given time by the force of gravity and the force urging 
the body down the plane, we have, section [4, or 113], these 
spaces proportional to the impelling forces. But these forces 
are, by the preceding article, proportional to the accelerations 
they are capable of producing in any given time — as the unit 
of time one second, for example. We have, therefore, 

S:S x v.g\f t 

or SiS.iig'.^j, 5:5 a ::i:~ S : S, :: I : h; 

which conforms to the enunciation of the proposition. It should 
be kept in view that in this, as in the examples following, the 
body is supposed to glide upon the plane without friction or 
other resistance. 

[124.] The time reqtdred for a body to f all throng] 1 the alti- 
tude of an inclined plane, will be to the time required for its descent 
along the plane, as the height of the plane is to its length. 

Let h and /, respectively, represent the height and length 
of the plane, T the time down the plane, and t the time of 



D YNAMICS. 



145 



falling through its altitude. We shall then have for the spaces 
accomplished in the assumed times 



i = ifr 



and 



h = \gt 



gh 



But /= ~- . Substituting this value of /, dividing one equa- 
tion by the other, and extracting the square root of each 
member, we obtain finally 



or 



i\ T:\ h:l, 



l_ T 

h ~~ t ' 
as enunciated. 

Corollary. — The times down any two planes of equal alti- 
tude will be as their lengths. 

[125.] The time required for a body to descend along the chord 
of a vertical circle, B A or A C ( Fig. 80), 
is equal to the time required for it to fall 
through the diameter B C. 

Resolving the weight of the body 
c b into its two components c a and 
a b, respectively parallel and perpen- 
dicular to the planes, and represent- 
ing the forces c a and c b by the ac- 
celerations they are capable of produc- 
ing in one second, denoted by f and g, 

we shall have from the similar triangles ABC and a b c> 
whose homologous sides are parallel, the proportions 

g : f :: be : ac : : B C : A C, for the chord A C, 

and g : / : : be : ac v. B C : A B, i or the chord A B. 

Denoting in either case the length of the plane by /, and 
the diameter of the circle by d, we have 




Whence 



:f::d: I. 
gl 



d- 



f 



146 ELEMENTS OF MECHANICS. 

Representing by T the time down the chord, and by t the 
time required to fall through the diameter B C y we have 

d=%gf and /=i/T\ 

Whence, by division, 

or replacing d by its value above, and reducing, we get 

Z g ? 

f-/ x r 

Whence T '=t, 

as stated in the proposition. 

The times of descending all chords drawn to the extremities of 
the diameter B C are therefore equal. 

Problem. — A point A lies without or within a given vertical 
circle. Through this point describe a second circumference, 
with A for its highest point, and tangent to the circumference 
of the given circle at a point P. Show that A P is the line of 
quickest descent from A to the circumference of the given cir- 
cle. This problem is a corollary to the preceding proposition. 

Let it also be shown that the line A P produced will pass 
through either the lowest or highest point in the given circle, 
according as the point A is without or within the circle. This 
being shown will give the following rule for finding the line 
of quickest descent : Draw a line through A and the lowest or 
highest point of the given circle as the case may require, and 
the line from A to the intersection with the circumference 
will be the required line. 

[126.] The velocity acquired by a body in descending an in- 
clined plane A B {Fig. 81) is equal to the velocity it ivould attain 
in falling through the altitude of the plane A D. 

Let V denote the velocity at B acquired in descending the 




DYNAMICS. 1 47 

plane A B, and V x that attained by the body in its fall from 
A to D. We then have, section [115], j 

V= V2JI and V x = VYgJ. 

Dividing one equation by the 
other, and substituting for /its value 

gh r n 1 

^j- [122], we have Fig. 81. 

V Vgk 

~ = --t==i, or VLV V 
K Vgh 

As the same could be shown for any other plane A C, it 
follows that the velocity accumulated upon planes of equal alti- 
tude is the same, and is independent of the inclination of the 
Planes. 

[127.] The initial velocities being the same, the time of descent 
upon the plane will still be to the time of falling through its alti- 
tude, as I is to h, as in section [124]. 

Let the initial velocity be denoted by a, and represent by 
Zand /, respectively, the time down the plane and the time 
of falling through its altitude (Fig. 81). We shall then have, 
section [118], 

A£ = l = aT+±fT 2 ; 

AD=h = at + igt\ 
Solving the first equation as a quadratic, we have 



a 



\/ 2 J+ a l = l * _ 



JF h 
or substituting for /its value -j-, we have 



T=— h {-a±V2gk + a*\, 

and from the second equation, in like manner, 



148 



ELEMENTS OF MECHANICS. 



o 



Hence 



h' 



or 



T: t :: l:h, 



as was to be shown. 

Corollary. — The time down any two inclined planes of equal 
altitude will be as the lengths of the planes. 

[128.] The initial velocities being the same, the final velocities 
will still be equal each to each, as in [126]. 

Calling Fand V x the final velocities, as in [126], we have 
[118] 

V=a+/T and V 1 = a+gt. 



But from the preceding section, 
T: t :: /: h, 



or 



gh 






Th 



also, / 

Hence 
V-V s =fT-gt= g -jT~ g ~T=o, 



or 



V= V x 



Corollary. — The velocity acquired by a body descending 
along the contiguous planes A G, G H, and 
H K (Fig. 82), leaving out of consideration fi 
the shocks sustained at the angular points, 
will be the same as that attained in falling 
through A D. For, by the article above, the 
velocities at B, C, and D are respectively 
equal to the velocities at G, H, and K. 

It is evident that when the planes be- 
come indefinite in number they will consti- 
tute a continuous curve, as A K, and the ve- 
locities at K and D will likewise be equal. fig. 82. 




DYNAMICS. 149 

[129.] If in a vertical circle (Fig. 80) two chords E C and 
D C be drawn to the extremity of the vertical diameter B C, the 
velocities acquired by a body in descending the chords will be as 
the lengths of the chords respectively. 

Let E C — I and D C = /„ and let V and V x be the corre- 
sponding velocities for the chords. Representing G C and 
EC by h and h v we have, section [126], 





V= \ f 2gh 


and 


K = 


\ / 2gh x . 


Hence 




V VI 
K ~ Vh, 






But from 


principles of Geometry we have 


EC 


*=z£CxCG 


and 


DC* 


= BCX CE y 


or 


l 2 = BCxh 


and 


A" = 


BC X K 


Whence 


r k 


or 


/_ 
l~ 


VJT 


Hence 




V I 

v x ~ 1; 







in conformity with the enunciation. 

[130.] The times of descent along planes unequally inclined 
to the horizon are directly as their lengths, and inversely as the 
square roots of their heights. 

Let / and /, be the lengths of two planes, whose altitudes 
are h and k Jt and let T and t be the times of descent. Denot- 
ing by / and /, the accelerations in one second upon the two 
planes, we shall have, section [113], 

l=ifT* and / f = i/ 1 /*. 



1 50 ELEMENTS OF MECHANICS. 



Hence 



T _ f 

T "4 



But ;/=*£ and /s J^ 



By substitution, — = — r— = -7—, 

or / : / : : -7= : -7=- 

Corollaries. 
If A = ^, 

we have T :/::/: /,. 
If, also, /, == //, 

we have 7^ : t ::l : k, as in [127]. 
If /=/ x , 

we have T: t :: Vh 1 : Vh. 

If the planes are parallel, then /'=*/„ and the original 
equations become 

f=i/T* and h = \ft\ 
or J: * :: 4//": V£. 

Under the last supposition, we have also, for the velocities 
acquired upon the two planes, 



V = V2fl and V\ = 4/27?,, 

or V: V, :: VT: Vh. 

[131.] In planes equally inclined to the horizon, the initial 
velocities being proportional to the square roots of the lengths of 



DYNAMICS. 151 

the given planes, the times of descent will likewise be in the same 
proportion. 

Recurring to [127], we have, for the time of descent of the 
plane whose length is /, the formula 



i/ 2 lf+a*-a 
f 

and for any other plane /, of equal inclination, 



V2l i f+a;-a 1 
f 

T V2I/+ a* -a 

Hence — = , — 

t i / 2 / i/+ a:-a 1 

But, by hypothesis, 

VT 

a : a.:: VT: VT„ or a, = a —~ • 

VI 

Substituting this value of a 1 in the last equation, multiply- 
ing both terms of the fraction by VT, and taking out from the 
denominator the factor Vl„ there results finally 

T__ (i/ 2 /f4-a*-a ) VT T_ VT 

t " ( V2 lf-\- ct -a) VT' * " ' Vt 

as enunciated. 

[132.] The same supposition continuing as in the preceding 
proposition, the final velocities will likewise be proportional to the 
square roots of the lengths of the given planes. 

Denoting by Fand V l the final velocities, h and h x the alti- 
tudes, and / and l x the lengths of the planes, as heretofore, we 
have, for the velocities at the expiration of the times Zand t 

[118], 

F- a +f T and V x = a, +ft, 

V a+fT 



I5 2 ELEMENTS OF MECHANICS. 

But, by the previous proposition, 

and by hypothesis, as in the same proposition, 

a = ay 7 . 

Substituting these values for T and a, and cancelling the 
factor (a, -\-ft), we have 

V__ VT 

k ~ vr; 

in conformity with the enunciation. 

[1 33.] The time of descent of a body along the contiguous 
planes I, / 2 , l v etc. (Fig. 83), will be to the time of descent along the 
equally inclined planes l x ,l 3 , / 5 , etc., included under the same angles, 
as the square roots of the sums of the tzvo sets of planes respectively, 
or as V7+/, + / 4 , etc., is to V l x + /, + / 6 , etc. 

It is evident, from principles of Geometry, that 

/: /, :: /, : /, :: l K : /., etc., 

and hence VT : VT, : : VT, '• VT 3 : : VT t : VT h , etc. 

But the velocities acquired upon the planes / and /„ which 
become the initial velocities to the planes . 
/ a and / 3 , are to each other [130, Cor.] 
as V~l to VT X , and hence as Vl~ to VT 3 . 
Hence, also [132], the final velocities at 
H and / are as VJ\ to VT 3 , or as V~h to 
VT b ; and finally the velocities at E and 
C are as VT t to f^. 

The initial velocities for the planes 
being, therefore, as the square roots of 
the lengths of the planes respectively, the times of descent 




DYNAMICS. 153 

will be likewise [131] proportional to the square roots of their 
lengths. 

Denoting these times by /, t v / 9 , t s , etc., we have, therefore, 

t : t x :: VT : VT V 

t, : t % :: VI: VT,:: VT : VT V 

tr-tr-'.Vl : VI:: VT : VT t ; 

or VT x \t + /,+ t< + etc.} = vTj/, + /, + /.+ etc.}, 

and V7: VT, :: (/ + /, + *« + etc.) : {t l +.t i .+ t % + etc.). 

But making (/ -f- £, + / 4 -f- etc.) ^ jT, the whole time of de- 
scent upon the planes /, / 2 , /„, etc., and (/, -f- t z -f- . /, -f- etc.) = 7^, 
the time upon the planes from i> to £T, we have 

1/7": */£:: 3T: r r 

But the first set of proportions gives 

/: i x : : (/ + / a + / 4 + e tc.) : (/, + / 3 + / 6 + -etc.). 
Hence 



T: T t :: Vl : Vl x :: Vl + /„ + / 4 + etc. : f/, + 7 3 -f / 5 + etc., 

<?.? enunciated. 

Corollary. — If the planes be supposed to become indefinite 
in number, and to constitute two similar curves D E and B C, 
we should still have 



T: T x :: VDE : VB C. 

[I34.] If a body A (Fig. 84) with an initial velocity a, descend 
an inclined plane A B in a given time t, and with its terminal 
velocity at B, as an initial velocity, ascend the equally inclined 
plane B C, for the same time, the spaces of descent and ascent will 
be equal. 



154 ELEMENTS OF MECHANICS. 

The space of descent will be given by the formula [118] 

and the final velocity at B, equal to (a -\-ft), 
becoming an initial velocity to B C = /„ 
we have by the same reference 




/, = (a +//) / - iff = at + \ft\ 
Hence / = l v 

as was to be shown. 

On the other hand, the spaces of ascent and descent being 
equal, the times will be equal ; for if the times were unequal, 
the spaces would be unequal, which contradicts the hypothesis. 

If, moreover, we denote by V the velocity at the expira- 
tion of the time t of the ascent, we have [118] 

V={a+ft)-ft = a, 

equal the initial velocity at A. 

Cor. i. — If a body descend from A (Fig. 85), along the inclined 
planes A B and B C, it will in the same time, with the terminal 
velocity at C as an initial velocity, ascend the equal and equally 
inclined planes CD and D E. 

For, by the preceding article, the time down B C with the 
initial velocity acquired at B will be 
equal to the time up CD; and since, by 
the same article, the velocity at D equals 
the velocity at B, the time up D E will 
likewise be equal to the time down A B. 
Hence the times of descent and ascent 
will be equal. 

Cor. 2. — The same method of reasoning, being applicable 
to any number of corresponding equally inclined planes, will 
also apply to the case in which the number of planes becomes 
infinite and constitutes a regular symmetrical curve. The 
times of ascent and descent upon such a curve would there- 




D YNAMICS. 155 

fore be equal whenever the two branches of the curve are 
similar and equal. This corollary will evidently apply to the 
case of a simple pendulum, as will be seen hereafter. 



Theory of the Pendulum. 

[135.] A simple pendulum may be conceived to consist of 
a ponderable material particle, attached to one extremity of a 
thread or inflexible line without weight, and oscillating in a 
vertical plane about a fixed point to which the other extremity 
is attached. 

The point about which the pendulum is supposed to vi- 
brate, is called the point of suspension. 

[136.] If a pendulum A B (Fig. 86) be moved to any distance 
from its natural vertical position AD, it zvill, if left free to fall 
without resistance, ascend to an equal distance D C on the opposite 
side, and the ascent from D to C zvill be in a time equal to that 
required in the descent from B to D. 

The truth of this proposition will be rendered apparent by 
reference to section [134, Cor. 2], as it may be shown to be the 
same in effect whether a body move 
without resistance upon the arc B C 
as a curve, or be confined by a cord 
A B so as to describe the same 
curve. For if we suppose the 
weight of the body to be resolved 
into its components, the one tan- 
gent and the other perpendicular 
to the curve at the point B, the component perpendicular to 
the curve would be equal to the reaction of the curve upon 
the body in the direction B A, which i 3 the same as if the body 
were held in the same direction by the equal tension of the 
cord A B. The tangential component tends equally in either 
case to produce motion or vibration. Hence the corollaries 




156 



ELEMENTS OF MECHANICS. 



i and 2 of section [134] are applicable to the case of the sim- 
ple pendulum. 

[137.] The velocities of two pendulums of equal lengths, at 
their lowest points, are to each other as the chords of the arcs of 
descent respectively. 

For the motion of the two pendulums A C and A E (Fig. 
86) would be similar to the motion of two bodies upon the 
curves CD and E D. But, section [128, Cor.], the velocities at 
D, the lowest point of the curve, would be the same as that 
accumulated upon the chords CD and ED, and these veloci- 
ties, section [129], are as the lengths of the chords respectively. 

[138.] The times required for two pendulums of different 
lengths to vibrate through arcs of the same number of degrees, are 
as the square roots of their lengths (Fig. 87). 

From section [133, Cor.] we have the time down BE is to the 
time down D G as VBF is to VDG; 
and since the times of ascending E C 
and G E are equal to the times of de- 
scending ^f and D G, section [134, 
Cor. 2], we shall have, denoting by T 
and T x the times of the whole vibrations, 




T: T x \\ VBE\ VUGwVbC: VDE. 



Fig. 87. 



But similar arcs being to each other as their radii, 
BC: DEw AB : AD, 



or VBC : VDE : : VAB : VA D, 

and T: T t :: VAB : VAD, 

as was to be shown. 

[139.] To fiyid a general expression for the time of vibration 
of a pendulum in a circular arc. 



DYNAMICS. 



157 



Let ab = A (Fig. 88) be an indefinitely small arc upon the 
curve of descent, and denote 
by / the time required for the 
pendulum to describe this arc, 
and by v the velocity at the 
point a, acquired in moving 
over Qa. 

By section [128, Cor.] this 
velocity is equal to that ac- 
quired by any body in falling 4 
through m e. 



Hence 

v 



V2J. 



m e. 



z 


\ 

m \ 


\y <r 


e \^ A / 


~\\ WNl 




*\ • 9 \ 


f J y 



But since the time in a b is FlG - 88 - 

indefinitely small, the velocity with which it is percurred 
may without error be considered as uniform; the variation 
in the velocity in a b being indefinitely small as compared to 
the finite velocity at a. 

A 



Hence 



v V 2g .m e 



Denoting the length of the pendulum by Z, we have, from 
similar triangles ab s and O a e, 

L.bs 



A : L : : b s : a e, 



or 



A = 



a e 



In like manner, from the triangles gc d and fc e, 

c d : c g :: c e : r, 
or b s : a : : c e : r, and 

and by substitution above, 

A 



b s±= —ce, 
r 



La c e 
r a e 



But 



c e 



= V 



m e.n e. 



and 



a e = yn e (2 L — n e). 



158 ELEMENTS OF MECHANICS. 



Whence 



r a \m e . n e I — t a 



r yn e (2 L — n e) V2g . m e r V 2 o-V 2 L — tie) 

or, multiplying both terms of the fraction by V2 and making 
;/ e = x, we have 

_L a V2 _ 1 J L a a/ 2 L 

2 r ' Vg V2L — x~ 2 ;g'r' 2L — x' 

Since a like expression could be found for the time on any 
number of very small arcs, upon the curve of descent, having 
corresponding arcs all equal a, upon the semi-circumference 
m n, the whole time of descent from Q to n would be equal to 
the sum of the elementary times similar to t. 

But if n x denote the number of arcs equal eg^=a in the 
semi-circumference m n, we have 

, a 7t 
n x a = n r, and — = — . 
r n x 

Hence, denoting by T x the whole time of descent from Q to 
71, we have 



g 1l x 2 L — X 

or the whole time of a full vibration equals 



g n x 2 L — x 

If now we suppose the arc of vibration to be very small, the 
different values of x in the expressions similar to y - 



2 L — x 

corresponding to n x different arcs, would all be very small in 
comparison with 2 L, and being neglected, there would be n x 

expressions equal to y —— = 1, or we should have 

2 L — X 



DYNAMICS. 159 

The equation for the time would then become 

g 

as a very close approximation for the time of vibration in small 
arcs, 

[140.] But the degree of this approximation to the exact 
time of vibration in small arcs may be more clearly seen thus : 

In the expression above we have, making OZ=On 
(Fig. 88), 



jf^L = 1 / 4 1? 



2L 2 L 2 L 



2L-x 2L(2L-x) Vjn.e~z ^7y * X* 

But z y = 2 L . cos n z y = 2 L cos -J- V, or, if a denote the 
whole amplitude of the vibration, z y = 2 L cos \ a. 
Hence, by substitution, 



T 



or 



g 


I 


2 


1 
COS J a 


g 


I 

n x 


2 


sec J ex. 



Now let the amplitude of the vibration be as great as 20 
even, and let there be 10 divisions of the arc on each side of the 
lowest point n of the curve. We should then have n x = 10 ; 
and if these arcs reckoned from ;/, and beginning with the 
largest, be denoted by a, a v a^ etc., the expression for T 
becomes 

T— rt\ — X — jsecJ«+ sec i a 'i + sec i <T 2 + et c. iotermsf, 

in which a — 20 , and a v a. v etc., are still less. 

But even if we should assume all of these arcs equal to a, 



l6o ELEMENTS OF MECHANICS. 

which would give a result greater than the exact time of 
vibration, we should have 

T = n V — • sec i a = it V — sec 5 , 
or T=«V£-X 1.00382 = * V- + -&■- «■ /-- 

g J £ ' IOOOOO ^ 

But it is obvious that 3 would be a much nearer approxima- 
tion. Making this supposition, 

7-=*/~X i.ooi 3 8 = ^/-- + -^- 7 rl/-. 

£\ 3 ^ ^ 100000 

So that under this supposition the error would be somewhat 
less than the y^W P ar t of the time given by the expression 

For the exact series giving the time of vibration in any 
arc, the student is referred to some work where the series is 
derived by the methods of the calculus. 

[141.] The time required for a body to fall through a space 
equal one half the length of the pendulum L, as derived from 
the formula \L = \gf, will be 



Hence the time of vibration of a pendulum through a very small 
arc is to the time required for a body to fall through \ its length 
as n to 1, or as 3.1416 to I. 

[142.] If there be two pendulums whose lengths are re- 
spectively L and L v and whose times of vibration at the same 
locality, through similar arcs, are T and T v we have 

T: T x :: VI: VZ~ y 
as in section [138]. 



DYNAMICS. l6l 

Hence it follows that if the time T of the oscillation of the 
pendulum whose length is L be known, the above proportion 
will give the length Z, of a pendulum corresponding to a 
given time of vibration T v or, on the other hand, will enable 
us to determine the time of vibration T x of a pendulum of 
known length L x . 

143.] In order to determine with precision the time of a 
single vibration, the following method may be adopted. Let 
/ be any assumed time during which the pendulum L per- 
forms N vibrations. Then, since the time of one vibration is 
equal to the whole time t divided by the number of vibra- 
tions, we have 

r-ir 

Similarly, if the pendulum Z, performs iV, vibrations in the 
same interval of time, we have 

These values substituted in the proportion above give 

-^ : ^ :: VL : VL V or N? : N>:: L : L lt 

and L, = — ^- 2 - • 

Knowing N, the number of vibrations of L in the time t, 
by experiment, it becomes easy to determine the length of 
the seconds pendulum for the latitude of the place of obser- 
vation ; for if iVj be assumed equal to the number of seconds 
in the time t, the corresponding value of L x will be that of the 
seconds pendulum. 

In the latitude of London this value of L x has been found 
to be equal to 39.13908 inches, and for New York 39.10153 
inches. 



1 62 ELEMENTS OF MECHANICS. 

[144.] If in the formula 



=«vz, 



for the time of vibration of the pendulum L v we suppose L l 
to become the length of the seconds pendulum for the lati- 
tude of London, g x being the acceleration due to gravity at 
the same latitude, then 



/ 



ft' 



or g x = 7T 2 L x — (3. 1416) 2 . (39. 1 3908 in.) =32.19 feet per second, 

for the acceleration in one second of the force of gravity at 
London, or the relative measure of the same force, section [7], 
the slight effect of the centrifugal force due to the earth's mo- 
tion (and to be considered hereafter) being disregarded. 

[145.] If we denote by g the acceleration of gravity in one 
second, at any other station, and by L the corresponding 
length of the seconds pendulum, we should in like manner 
have 

g = 7T 2 L. 

But if G x and G denote the intensities of the force of gravity 
at the two stations, we have 

G • G '- 2 ■ F • • L • L 

Hence it appears that tJie intensities of the force of gravity at 
any two stations of different latitude are proportional to the lengths 
of the seconds pendulum for the stations. 

[146.] Recurring to the proportion 

1 n- sr ' 



let T and T x represent the times of vibration of the j 
dulum for two different latitudes. Then L = L v and 



same pen- 



DYNAMICS. 163 

or, replacing Tand T, by -^ and-^, in which N and N x de- 
note the number of vibrations in the time t, at the two places 
of observation, we have, after reduction, 

Nr.N':: gi :g, or g = g^- 

If g denote the force of gravity at the equator, as deter- 
mined by the method already given [144], and iV^the number 
of vibrations of the given pendulum in any assumed time, the 
last equation would give g v the intensity of the force of grav- 
ity for any other latitude, where the number of vibrations N x 
could be determined by experiment. 

Thus, the force of gravity at the equator being ascertained, 
and N the number of vibrations of a given pendulum in a 
given time, we are enabled by a simple experiment with the 
same or an equal pendulum at other stations to determine the 
force of gravity for any given latitude. 

[147.] Let it be required to ascertain the effect of lengthening 
or shortening the seconds pendulum a given small amount, upon 
the number of vibrations in any given period, say one day. 

If, in the equation 

we suppose L to become equal to (L 1 -\-p), the seconds pen- 
dulum increased by p, and N to become (iV, — q), where q 
denotes the number of vibrations by which N r is diminished 
in a day, in consequence of the increase of L x we have 



{N t -qf -N?~2-N t q+f 



m+* L >wj 



164 



ELEMENTS OF MECHANICS. 



But if the loss in the number of vibrations be small, amount- 
ing only to a few seconds of time, q will be very small in com- 
parison with N v the number of seconds in one day, the period 

chosen, and consequently l~~) being a very small fraction, its 

higher powers may be neglected, and there results 



L l +p = L l + 2L i 



or 



P = 2 A 



M 



and 



n: 



? = 



N x p 

2L: 



The first equation shows the increase of length / due to a 
given loss q in the number of vibrations, and the second the 
loss q in the number of vibrations, due to a given increase of 
length p in the seconds pendulum. 

[14-8.] To find an expression for the time of vibratio7i of a 
simple pendulum in a cycloidal arc (Fig. 89). 

Let B G D be the generating circle, and denote its diameter 
BVbyd. 

Let PQ be a cycloidal cheek similar to BR, and Pa a 
very small and flexible cord attached to a material particle 
at a. 

By a property of the cycloid, when P V is equal to VB f 
the particle a will vibrate in 
the cycloidal arc Q B R. Put 
s z = x, and represent by L 
the length of the pendulum 
PB — 2d. Let also ny = r. 

From the manner in which 
the cycloid is generated, the 
chord K H is obviously per- 
pendicular to the tangent to 
the curve at the point H. 
Hence H L, drawn to the ex- 
tremity L of the diameter of the generating circle in the posi- 
tion given, is tangent to the curve at H. 




DYNAMICS. 165 

But H L is parallel to the chord B G. In like manner it 
follows that the tangent a T at the point a is parallel to the 
chord BD. 

Hence, assuming the arc a b to be indefinitely small, the 
triangle a b m will be similar to B D Z, and we shall have 

ab : dm :: BD : BZ\ 



and since B D — VB V . B Z, and b m — no, 

, . VBV.BZ v ,/5T 

we have a b = --— X no = noy —-. 

h Z, Jt> z 

But from similar triangles, 

« : np w n z \ ny, 



nz x V5Z.^Z 

or « = «/ • — = #/ . 

71 y ny 

Hence 



VSZ.BZ JBV ^VBV.SZ ^Vd.x 
ny BZ 71 y r 

But since the time required to describe the elementary 
arc a by sections [6 and in], is equal to 

ab 



V 2gX 

we have by substituting for a b its value, 

Vd np 

Hence it follows that the whole time from A to B, equal to the 
sum of the elementary times, and equal to i the whole time 
of vibration T, will be 

Vzg r 2g- r 2g 



1 66 



ELEMENTS OF MECHANICS. 



and the whole time of vibration will be equal to 
T. 



2 £ g 



This expression being for the time independent of the arc 
of vibration, it follows that all the vibrations of the cycloidal 
pendulum, whatever be the amplitude, are isochronous. 

Cor. — If we denote by / the time required for a body to 
fall freely through the diameter d of the generating circle, 
we have 

d=\gf and t-V— = i/.- 



L 



or, the ti7ne of any complete vibration is to the time required for a 
body to x all through the diameter of the generating circle as n to I. 



Central Forces. 

OF THE MOTIONS OF BODIES AROUND A CENTRAL FORCE. 

[14-9.] If a body revolve about any CENTRAL FORCE, the line 
joining this centre with the revolving body, called the RADIUS- 
VECTOR, will sweep over equal areas in equal times, or areas pro- 
portional to the times. 

When applied to the motions of the planets about the sun 
as a centre of attraction, this be- 
comes the enunciation of the first 
law of Kepler. The truth of the 
proposition may be thus estab- 
lished. Conceive a body b (Fig. 
90) to move, under the action of 
some initial impulse, from b to p, in 
some assumed unit of time. It is 
obvious that, under the supposi- 




Fig. 90. 



tion that there is no resisting medium, it would in the second 



DYNAMICS. l6y 

unit of time move through p m = bp. Let us suppose, how- 
ever, that when it arrives at/ it receives an impulse from a 
central force situated at C sufficient to bring it, if free to move, 
to a, in the assumed unit of time. Thus the body at/, acted 
upon simultaneously by two forces, the one tending to carry 
it to m and the other to a in the same time, would, by the 
principle of the parallelogram of forces, move through / n, the 
diagonal of the parallelogram a m, and arrive at ;/, at the ex- 
piration of the second unit of time. 

Having arrived at the point n, through the diagonal/;^, it 
would, if unaffected by the central force, move in the third 
unit of time through no = p n. If, however, when it arrives 
at 71, it receives an impulse from the central force sufficient to 
bring it, if free to move, to v, in the chosen unit of time, it 
will, as before, move through the diagonal n s of the parallelo- 
gram vo, and reach s at the expiration of the third unit of 
time. The respective areas described by the radius-vector in 
the first, second, and third units of time will be the triangles 
cbp, cpn, and ens. Joining cm and c o, we have, since the 
triangles c bp and cp m have equal bases, bp and p m, and the 
same altitude, c bp = cp m. Also cp m and cp n, with same 
base cp and equal altitudes, are equal. Or 

cbp = cp m = c p 7t. 
In like manner, cp n = c 71 o = c us. 

Hence cbp = cp7i = c7i s f 

or the areas described by the radius-vector, in successive equal 
units of time, are equal ; and the areas generally are, there- 
fore, proportional to the times. If we now conceive the inter- 
vals of time between the impulses of the central force to 
become infinitesimal, the path of the body, instead of being a 
broken line, would become a continuous curve, and the cen- 
tral force a constantly acting pressure, varying or not in in- 
tensity as the case might be. It is obvious that the proof 



1 68 



ELEMENTS OF MECHANICS. 



which has been given, will not be affected by our supposition 
as to the intervals between the impulses of the central force, 
but will hold true to the limit. 

It appears, therefore, that in whatever path a body may 
revolve through an unresisting medium, when subject to the 
action of a deflective central force subsequent to an initial 
projectile force, the law of areas will obtain. This is called 
the general principle of equal areas ; and very little considera- 
tion will also show, that if the central force be supposed to be 
repulsive instead of attractive, the same principle will hold. 



OF MOTIONS IN CIRCULAR ORBITS. 

[150.] When the centre of force is at the centre of the circle 
(Fig. 91). 

Since the areas swept over by the radius in equal times 
are equal [149], and equal areas 
are included under equal arcs, it 
follows that equal arcs are de- 
scribed in equal times, and hence 
the revolutions are performed 
with a uniform velocity. It is 
also evident that for equal arcs 
a b and c d, the deflections from 
the tangents ;/ b and ;;/ d are 
equal. But the amounts of de- 
flection at any two points in the 
orbit, for an indefinitely small 
time, being the same, the intensity of the central force at these 
two points must be the same, and we conclude, that in circular 
orbits, the centre of force being at the centre of the circle, the in- 
tensity of the force remains constant. 

If we conceive the revolving body b to be connected with 
C by a small cord or string, the amount of tension of the cord 
would represent in magnitude either the central or centrifu- 
gal force, these two forces being the same in value, but reck- 
oned in opposite directions. 




Fig. 91. 



DYNAMICS. 169 

[151.] The acceleration which the central force, acting with 
uniform intensity for one second, would impart to the body, is equal 
to the square of the velocity of revolution, divided by the radius 
of the circle. 

Take the arc a b (Fig. 91) indefinitely small. Let/ denote 
the acceleration the central force is capable of imparting in 
one second, v the velocity of revolution, t the time of describ- 
ing the arc a b, and r the radius of the orbit. 

Since a b is indefinitely small, the direction of the central 
force for the time / may be supposed to remain parallel to 
C a, and the chord ab to be at the limit equal to the arc ab. 
It belongs to geometry to show that the difference between 
the chord and arc at the limit is infinitely small in comparison 
with either chord or arc, and may therefore without error be 
neglected. 

We also have from principles of geometry a$ = 2 r . ag, 

— a~b 2 
or ag = 

2r 

But as the velocity of revolution is uniform, section [150], we 
have 

a b •=. v t. 

7 ,2 f 

Hence ag= - — • 

a 2 r 

But since ag = ~b~n is the deflection, or space through which 
the central force as a constant force moves the body in the 
instant of time t, we have [113] 

v* i 
Hence \ff — 



2 ^2 



2 r 



and/, the acceleration due to the central force for one second, 
will be 



170 ELEMENTS OF MECHANICS. 

in which equation it should be observed that if /and v are 
measured in feet per second, r must also be measured m feet, 
and so for any other assumed linear unit of measure. 

The magnitude of a constant force being referred to the 
acceleration it is capable of imparting to a given mass in one 
second, as a relative standard of measure, with other like 
forces, this value of /will therefore represent the intensity of 
the centripetal or centrifugal force, in comparison with other 
forces of the same kind, acting upon the same mass, and pro- 
ducing a different acceleration. That is, the forces will be in 
the ratio of the accelerations due to them. 

Corollary 1. — The velocity of revolution will be equal to that 
which the central force, acting with uniform intensity, zvould im- 
part in accelerating the body through \ the radius. 

From the equation f=- we have v = Vfr, /being the 

r 

acceleration the central force can impart to the body in one 

second. But the velocity acquired by a body moving under 

the action of a constant force, whose acceleration for one sec- 

oudisf through a space equal to i r, will be [113] 



v x = V~2fs — V 2f- \r = Vfr. 
Hence v x — v, in conformity with the enunciation. 

Cor. 2. — Let tzvo bodies of equal mass revolve in different cir- 
cular orbits. 

Denote by F and F x the intensities of the centripetal 
forces, by v and v x the orbital velocities, and by r and r x the 
radii. Then, since the two forces are proportional to the 
accelerations they are capable of imparting in the unit of time, 
we have 

F:F x v.- :^- 

r r x 

It may be well to remark here, that, as in the case of the force 
of gravity, the centripetal force varies directly as the mass, 



DYNAMICS. 171 

and hence the acceleration due to it is independent of the 
mass, in any given orbit with a given velocity. 

(a) If the orbits are equal, or r = r v 

F: F x ;\ v 2 : v x \ 

and the central forces are as the squares of the velocity. 

(b) If the velocities of revolution are equal, 

F: F x :: r x : r, 

or the central forces are inversely as the radii. 

(c) If the times of revolution, or periodic times, which we may 
denote by T and T x , be introduced, we shall have, since the orbital 
motioyi is uniform, 

~ 2n r , ~ 2n r. 
1 — and 7, — -, 



v v x 

or v 2 = ^ and v x = - lL y F 1 



Hence F. F, .. —- : -— .. - . - 



If T — 7i, we have F : F x :i r : r x . 

(d) If T* : T 2 :: r z : r x \ whence T 2 = ^-f-, we have 



h • h • • ^— * — - * * — * — 



that is to say, when the squares of the periodic times are propor- 
tional to the cubes of the radii, the central forces will be inversely 
as the squares of the radii. 

This corresponds to the third law of Kepler, to be dis- 
cussed hereafter. 

Cor. 3.— It has been shown that if any body revolve in a 
circular orbit of radius r, the central force F necessary to 
maintain the circle would, if it acted as a constant force upon 



1 72 ELEMENTS OF MECHANICS. 

the body, moving it in a rectilinear direction, impart to it in 

one second an acceleration equal to — . 

r 

If, now, we suppose the weight of the body to be W, and 

the acceleration due to gravity, imparting this weight, to be 

equal to g, we shall have, since constant forces are as the 

accelerations they will impart to the same mass in the same 

time, 

F : W :: - : g, or F — . 

r g r 

Since g and Ware known quantities, varying only with the 
latitude, and v and r are supposed to be given, the intensity 
of the central or centrifugal force F becomes known in terms 
of the unit of weight adopted. 

Since, also, the mass M of any body is represented by the 

• w 

constant ratio — , we have 
g 

_ Mi? 
F= , 



a formula for the centrifugal force in terms of the mass of the 
revolving body, the orbital velocity, and the radius of the 
orbit ; r and v being measured in the unit of g. 

Cor. 4. — If we denote by N the number of revolutions, or 
the part of a revolution, described in one second, we shall 
have 





space 2 71 r X N 

v = - = = 2 7t r .N 

time 1 


and 


v* = 4 ttV . N\ 


Whence 


F= W.r.N^— = 1.227 Wr.N\ 



a formula for the centrifugal force in terms of the number of 
revolutions per second, the value of g used being always that 



DYNAMICS. 173 

corresponding- to the latitude giving the weight W\ or the 
formula may be written 

F= 4tt\ rMN\ ■ 

{Examples for solution to be supplied by the Professor?) 

OF THE MOTIONS OF BODIES IN VARIOUS ORBITS. 

[152.] To determine the relation between the velocities at any 
points of an orbit, and the perpendiculars upon the tangents at 
those points. 

Let A B (Fig. 92) be the given orbit, v and v x the veloci- 
ties at P and P lf and p and /, ? L 
the perpendiculars upon the ^r^^^v^ 1 
tangents. Let A and A x be 
the areas swept over by the / 
radii - vectores in the indefi- 
nitely small time t. The arcs 
described respectively in the 
given time, since the motion Fig. 92. 
for an indefinitely small time may be regarded as uniform, 
will be 

vt and v 1 t. 

But these arcs are the bases of the triangular areas de- 
scribed by the radii-vectores, and the perpendiculars upon the 
tangents are the altitudes. 

p p 

Hence A—-.vt y A, = £ ^.v 1 t; 

or, since the times being equal, the areas are likewise [149], 
we have 

v A 
vp = v,p v or v x — J> 

that is, the velocities at different points in any orbit are to each 
other inversely as the perpendicidars upon the tangents. 




1/4 ELEMENTS OE MECHANICS. 

Cor. — From the last equation we have 

vp = v t p 19 

and the product of the velocity at any point in the orbit, by the per- 
pendicular upon the tangent at that point, is a constant quantity. 

Oscillatory Circle (Definition). 

If P be a point on any curve, a circle whose centre is on a 
line perpendicular, on the concave side, to the given curve at 
P, and whose circumference passes through this point, may 
have greater or less curvature at that point than the given 
curve. Hence between the circumferences of greater and 
less curvature there will be some one circumference which, 
at and near the point P, will coincide most intimately with 
the given curve. This is called the oscillatory or equicurved 
circle. 

[153.] To determine a general expression for the lazv of the 
centripetal force in any orbit. 

This expression may be found first in terms of the radius- 
vector, perpendicular upon the tangent, and the radius of the oscil- 
latory or equicurve circle. 

Let C (Fig. 93) be the centre of force, A B the orbit, 
CP= r the radius-vector, C 't = p the perpendicular upon the 
tangent, and PR = R the radius of curvature at the point 
P. Let Pm = F represent 
the intensity of the central 
force upon the body at P. 
Now, whatever be the nature 
of this force, it may be re- 
solved into the two compo- 
nents Pn —f in the direction 
of the centre R of the oscilla- 
tory circle at P, and Pq in 
the direction of the common FlG ' 93 ' 

tangent to the circle and orbit. But at and near the point P 
the circle and orbital curve coincide, and the bodv for an in- 




DYNAMICS. 175 

stant moves in both. Moreover, from the law of motion in 
circular orbits, section [151, Cor. 3], we have, denoting the 
mass of the revolving body by M and its velocity by v, 

Pn =/: 



R 

And similar triangles Pm n and P C t give 
Pm : Pn :: PC: Ct, 
or F : f :: r : p. 

Hence F=-f=Mv'-~) 

P Rp 

or, multiplying numerator and denominator by p 2 , 

and since, by section [152, Cor.], v 2 f is constant, the central 

force varies as the expression -— - . 
v Rp> 

[154.] In order, therefore, to determine the relation be- 
tween the central force and the radius-vector, or distance of 
the body from the centre of force, it will become necessary, 
in the case of any particular orbit or curve, to find an expres- 
sion for the value of Rp % in terms of r ; which, when substi- 
tuted in the preceding equation, will give the value of the 
central force F, in terms of r, the distance of the revolving 
body, and constants. The derivation of this value of Rp* for 
curves in general, is more readily accomplished by the meth- 
ods of the differential calculus. As it is important, however, 
to discuss the laws of the motions of bodies in elliptic orbits, 
because of their interesting application to the bodies of the 
solar system, a method of determining the value of Rp 3 in the 
case of the ellipse, the locus of the force being at one of the 
foci, will be found in Appendix A at the end of the volume. 
The solution for the hyperbola and parabola will also be found 
under the same reference. 



176 ELEMENTS OF MECHANICS. 

[155.] As lound in this reference to the Appendix, we 

have, denoting the semi-axes of the curves by a and b, and the 

semi-latus rectum by s, 

b 2 
For the ellipse, Rp z = r*-=r*s; 

a 

b 2 
For the hyperbola, Rp 3 = r z — =r*s; 

a 

For the parabola, Rp* = r z s. 

Substituting- these values of Rp 2 , there results, for the 

ellipse and hyperbola, F — Mv 2 p 2 - . -, 

and for the parabola, F = — — — • — - 

s r 

Hence, since v 2 p 2 is a constant value, section [152, Cor.], it 
follows that, when the orbit is one of the conic sections, the locus 
of the central force being at one of the foci, this force will vary 
inversely as the square of the distance of the revolving body. 

[156.] Since vp is equal to double the area described by 
the radius-vector in the unit of time, if we denote this area by 
A, we have vp = 2 A and v* p 2 = 4 A* ; and if e be the eccen- 
tricity of the orbits, then in the case of the ellipse 

2 tf . a 1 

1 — e = --, whence 



a 2 ' b 2 a(i-ej 

and for the hyperbola 



2 b 2 a 

e = r, or 



b 2 a (e 2 - 1) 

These values for v 2 p 2 and -, substituted in the formula of 
the preceding section, give, 

For the ellipse, F — -4 7- • - ; 

a (1 — e ) r 

For the hyperbola, F = — — — . - ; 

Jr a {/ — i) 7" 



DYNAMICS. 177 

For the parabola, F = . - ; 

s r 

and also for the circle, since e = o, 

u ^ MA " l 
F=- . -. 

a r 

In the case of the circle, since for the same circle r and 

a = r are both constant, as well as A, the force F remains con- 

r 
stant, as should be the case. Again, since A = v X 1 X — , we 

have 4 MA* = Mzfr*, which substituted gives 

_ Mi?r* _ Mv* 

~ r 3 ~ r 

as has been already found [151, Cor. 3]. 



THE ELLIPTIC ORBIT. 

[157.] If a body revolve in an elliptic orbit, the centre of 
force being at the focus, then will the velocity at P (Fig. 94), 
the extremity of the minor axis, be just sufficient to maintain the 
body in a circular orbit, around the same central force, with a 
radius equal a, the mean distance in the ellipse. 

Let R be the radius of curvature at the point P, equal 



to —j- (Appendix), and let F de- 
note the central force, and / the 
component of the same force in 
the direction of C, the centre of 
the equicurve circle at P. The 
force / is obviously that which 
would maintain the body in the 
osculatory circle at and very near 
the point P, and hence, section 
[151, Cor. 3], we have 

_ Mv i 





Y- 


p 




f / ' 




\ 
\ 


\ 


/ 


/ 
/ 
/ 
/ 

S 

s 


c 


1 


* 
^ 


^--_L_- 


■—-""" 




Fie 


1. 94. 





178 ELEMENTS OF MECHANICS. 

in which v is the velocity in both the circle and ellipse at the 
point P. But the central force F being resolved into / and a 
tangential component Pn, we shall obviously have, from simi- 
lar triangles, 

or f=F-> 



F: 


:f::a 


:A 


a 


Mv* 

" R ' 





Hence F— = „ — = — and F = -, 

a R a a 

which is seen to be the force required to maintain a circle, 
with the velocity v and radius a, as was to be shown. 

158.] Moreover, the time of revolution, or periodic time, in the 
ellipse, will be equal to the periodic time in the circle. 

Since in any orbit the areas described by the radius-vector 
are proportional to the times, if we divide the whole area of 
the ellipse or circle by the area swept over by the radius-vector 
in the unit of time, the result will give the periodic time for the 
orbit. Let Ps and Ps 1 (Fig. 94) be the spaces that would be 
percurred in the circle or ellipse with the velocity at P, equal 
to v, considered as uniform, in the indefinitely small time /. 
Then we shall have Ps = Ps l = v t, and the triangular areas 
PFs and PFs v described by the common radius-vector a, will 
be equal respectively to the bases of the triangle, by one half 
their altitudes ; that is, we shall have 

PFs = vt .— and PFs, = vt.~; 
2 l 2 ' 

and if t be assumed as a very small unit of time, 

PFs — v- and PFs, = v-' 

2 * 2 

Hence, denoting by T and T x respectively the periodic times in 
the circle and ellipse, we shall then have, since the area of the 
circle is n a 2 , and of the ellipse nab, 

n a 2 2 7t a _ _ nab 2n a 

1 = = and T. = — ,- = , 

a v a b v ' 

v- v — 

2 2 

or T= T v as was required to be shown. 



DYNAMICS. 179 

[159.] The velocity v of the revolving body, at the point P (Fig. 
94), will be equal to that which it would acquire if allowed to fall 
freely, under the action of the central force at that point, consid- 
ered as constant, through one half the mean distance a. 

For the velocity v at the point P has been shown [157] 
to be equal to the velocity which would maintain a circle with 
a radius equal to a, the mean distance in the ellipse. But [151, 
Cor. 1] this velocity is equal to that attained in falling through 
one half the radius, or in this case one half the mean distance 
in the ellipse. Hence the truth of the proposition. 

[160.] To find an expressio?i showing the relation between the 
squares of the periodic times, and the cubes of the 7nean distances, 
in two different orbits about the same central force. 

Let a and b be the semi-axes of one of the orbits, T the 
periodic time, M the mass of the revolving body, and r its 
distance at any moment from the centre of force at the focus. 
Let also A denote the area described in the unit of time ; then, 
since n a b is the area of the ellipse, we have 

_ 71 a b 
~~A~' 



But (1-^) = ^, or b = aVi 



Hence T 



or A = 



TtaWi 



A 
rt a 2 VT~~e 2 



T 
_4Mn 2 a*(\ — e 2 ) 



and 4MA 

Substituting this value of 4 MA 2 in the equation 

F = 4 MA 2 i_ 
~ a(i -e 2 ) V 



l8o ELEMENTS OE MECHANICS. 

section [156], we have 

- r , . ^ ; 

and in like manner for the second orbit we should have 

_ 4 M tit a? 1 . 

1- — y*— •--»; 



F_ 
whence ~~~a i ~T*r T 



M, 



Now, first, if the squares of the periodic times are propor- 
tional to the cubes of the mean distances, then we have 

a 3 t; F ■ & 1 1 

But — and -~ are the intensities of the central force upon 

unit masses of the two bodies, and are measures of these two 
forces. 

Hence we conclude that, if the squares of the periodic times 
in the two orbits are proportional to the cubes of the meati dis- 
tances, the central forces will vary inversely as the squares of the 
distances. 

Conversely, it is obvious, from the same equation, that, if 
the central forces are inversely as the squares of the distances, the 
squares of the periodic times will be proportional to the cubes of the 
mean distances. 

Thus it is seen that these two propositions are reciprocally 
dependent upon each other. The latter — viz., that the squares 
of the periodic times are proportional to the cubes of the mean 
distances — was first discovered by Kepler to be true of the 
bodies of the solar system by observation ; and it is from the 
above seen to be a consequence of the law of gravitation. 
Before the discovery of the law of gravitation, as applicable 



DYNAMICS. l8l 

to the planetary bodies of the solar system, by Newton, the 
following laws relating to the motions of these bodies had 
been ascertained by observation by Kepler, and hence have 
been called Kepler's laws. They may be thus stated : 

1 . The radius-vector of every planet describes about the sun as a 

centre equal areas in equal times. 

2. The paths of the planets are ellipses with the sun at one of 

the foci. 

3. The squares of the periodic times are as the cubes of their 

mean distances from the sun, the centre of the system. 

In the foregoing sections, we have shown that the first law 
would be true, whatever might be the orbit or the law of the 
central force ; that from the second it would follow that gravi- 
tation in any specific orbit would vary inversely as the square 
of the distance. But to establish the universality of the law of 
the force from one orbit to another, let k denote the force of at- 
traction at the unit's distance, upon the unit mass in any orbit, 
and A the area swept over in the unit of time by the radius- 

b f~k 
vector. It will be shown [170] that A — - V — ,a and b being 

the semi-axes of the orbit. Hence if T denote the periodic 
time, we have 



nab ZTid* 4 * 2 a' 
J ~ A - iTF> ~ k ' 


T 2 
a 3 


= ~T~ : 


and for any other orbit, 






77 4 7t* 







Hence, assuming the truth of the third law of Kepler, as 
shown above, we have 

whence not only is the law of the force the same, but the abso- 
lute force is the same, and if the planets were at the same dis- 



1 82 ELEMENTS OF MECHANICS. 

tance from the sun, the unit of mass in each would be equally 
attracted to the centre. 

Note. — Modification of the Third Law of Kepler. — In the fore- 

going expressions for the values of — and -- 1 , denoting the 

accelerations the central force could impart to the two revolv- 
ing bodies M and M v this central force, which we may regard 
as a force of attraction, has been supposed to act upon masses 
which possessed the property of inertia alone, and no proper- 
ty of reciprocal attraction for the central body. In fact, how- 
ever, the planets exert an attractive force upon the sun pro- 
portional to their masses, and induce an acceleration in the 
sun equal to the acceleration they would induce in masses 
equal to themselves placed at the sun's distance, just as the 
force of terrestrial gravity imparts the same velocity to differ- 
ent masses. The additional tendency of the planet to the cen- 
tral sun would then be equal to that due to an increase of the 
mass of the sun by the mass of the planet, and the additional 
acceleration of the planet toward the central sun would be 
the same as the acceleration the planet could induce in the 
sun. Let 5 represent the sun, and M the planet. The attrac- 
tion between them, section [8], will be, at the distance r, 

SM 



and the accelerations of the sun and planet respectively, sec- 
tion [4], 

SM . ■ M A SM ._ 5 

— — -r-S = - T and — — -r-M= — 
r 1 r* r* r 2 ' 

Hence the total acceleration of the planet toward the sun by 
the reciprocal attraction, equal to the sum of these accelera- 
tions, will be 

S + M 



DYNAMICS. 

and for another planet, M v 

S+M, 



183 



27 

Substituting these values in the preceding equations for — 

and -~r. we have 
M, 

S + M 

S + M, a? T* A r 2 ' 



whence 






(S+M\ 
\S4-mJ' 



the 'exact expression for the law as applied to the planetary 

bodies of the solar system. But the fraction ( c ~f n/r \ owing 

to the great mass of the sun as compared with any of the 
planets, approximates very closely to unity. Let the student 
substitute the quantities S, M, and M, for the sun and any 
two planets, whose masses relatively to the earth are given in 
most works on astronomy, and the degree of this approxima- 
tion will be made manifest. 

[161.] The velocity v at P, the extremity of the minor axis of 
the ellipse (Fig. 95), is a mean proportional to the velocities v x and 
v^ at equal distances from P. 

For, section [152], Up be the perpendicular upon the tan- 
gent at v v we shall have, 
since the perpendicular upon 
the tangent at P is b, 



t 



or 



b* 



But by a well-known prop- 
erty of the ellipse, if perpen- 
diculars p and p x be drawn 
to the same tangent from 




FiG.95. 



1 84 ELEMENTS OF MECHANICS. 

the foci/ and f v their product will be equal to the square of 
the semi-minor axis, or pp x — b\ Hence 

v 2 _ f_ _p 

v';~Jp\~a 

Now if a tangent be drawn at v v it is obvious that/ would 
become/,, and/, would become/. 
Hence we should have 

<- p> 

or, by multiplying together the last two equations, 



= i , v* = v x v v and v = 1^ z/ a 



as was required to be shown. 

s . p r 

Cor, — Since — = — , we may also write 
P\ f\ 

v r 

< = r, ' 

and if v % be the velocity at any other point in the orbit, at 
which r and r, become r 2 and r 3 , we shall in like manner have 



v 
Z7% 



Whence -'- = -'A 

[162.] Since at any two points in the orbit the velocities 
are inversely as the perpendiculars upon the tangents, it fol- 
lows that at any two points of the curve equidistant from 
either extremity of the transverse axis, where these perpen- 
diculars are equal, the velocities will also be equal. 

Again, if v x and v 2 denote the velocities at A and B, the 



DYNAMICS. 185 

perpendiculars upon the tangents will be respectively (a -\~ a e) 
and (a — a e), and hence we have 

v x _a — ae _\ — e 

v 2 a-\- ae 1 -\- e 

If e = o, the ellipse becomes a circle, and v { = v v as should be 
the case. 



THE SPECIES OF THE ORBIT. 

[163.] To determine the relations connecting the values of the 
central force, velocity, and radius-vector, at any particular point of 
the orbit, that will determine its species. 

Resuming the equation F= — — -- — . -, section [155], and 

denoting by / the acceleration of the unit mass in the unit of 
time, we have, section [7], 

*L = f -*? Pi 

M J r* b* 

But if p x be the perpendicular from the second focus, upon the 
tangent at the point of the orbit whose distance is r from the 
centre of force, we have, from the properties of the curves, 

/A = b\ 
and hence 



b* A 



r. 



But in the ellipse and its limiting curves, the parabola and 
circle, 

r r 



r x (2 a — r) 1 
and in the hyperbola, 

r r 

r, (r -f- 2 a)' 



1 86 ELEMENTS OF MECHANICS. 

These values of r — give by substitution 

- v 2 a i v 1 a 



r (2 a — r) 2 a r — r % 
for the ellipse, parabola, and circle, and 

-,.2 ~ 

/= 



r -\-2ar 
for the hyperbola. 

Deriving the value of a from these equations, we have 
from the first 

_ /*> 



and from the second 



*-2/ r --V 



fr 2 
v — 2fr 



b 2 b 
But in the first three curves 1 -? e* = — , or a = — , and 



a 2 ' |/i _ 



JL2 Jl 

in the hyperbola e 2 — 1 = — , and # = 



« tV - 1 

Hence for the ellipse, parabola, and circle, 

_b__ fr 2 

VT~-^~7~ 2fr-v 2 ' 
and for the hyperbola, 

b fr 2 

V7^1 ~ v 2 - 2fr 

Now for the ellipse, parabola, circle, and hyperbola, respec- 
tively, we have 

e < 1, e = 1, e = o, e > 1. 

If e < 1, then — is positive, and hence v 2 < 2 />. 

Ellipse. 



DYNAMICS. 1 8/ 

If e — i, then g i s equal to infinity, or v 2 = %fr~ 

Parabola. 



If e = o, we have £ = — / -. But b = r, and 

2/r — v 2 



we 



have 1 = —/ -, which gives 2fr — v 2 =fr, or v 2 = fr. 

2 f r — v 

Circle. 

In the second equation — - for e > 1 is positive, and 

V e 2 — 1 

thence we have for the hyperbola the condition v 2 > zfr. 
To resume, the equations of condition are, 

For the ellipse, v 2 < ifr ; 

For the parabola, v 2 = 2 fr ; 

For the «Vr/<?, ^ 2 =fr, as in section [151] ; 

For the hyperbola, v 2 > 2fr. 

There are two singular results involved in the above dis- 
cussion that deserve the careful attention of the student. It 
will be observed that the formulse just derived, determining 
the species of the curve or conic section, as well as the formula 
giving the value of a, the semi-major axis of the orbit, have 
been deduced without any reference to the value of the angle 
which the direction of the motion of the body makes with the 
radius-vector, and this angle does not enter into these equa- 
tions. In other words, the species of the orbit, as well as the 
length of the major axis of the curve, is entirely independent of 
the direction of the initial impulse, which would cause any projected 
body to take up its course around a central force, varying inversely 
as the square of the distance. 

This of course excludes the case in which the body ma)< 
be conceived to be projected directly toward the centre, 
when there could be no deflections into a curve, in which 
case the orbit would become developed into a straight 
line. 



1 88 ELEMENTS OF MECHANICS. 

[164.] To introdtice the angle made by the direction of the 
motion with the radius-vector into the discussion. 

Resuming the equation of section [156] for the ellipse, 

4MA* 1 4 A* 1 



a (1 -e*) r 2 ' ^ J ~ (1 - e*)r 2 a' 
we have, section [156], 

A=^ and ±A* = v 2 p\ 

But it is evident that if a denote the angle between the radius- 
vector and the direction of the motion, or tangent to the 
curve, we shall have 

p = r sin a and v*p 2 = 1? r 2 sin 3 a = 4. A*. 

Whence, by substitution, 

v 2, sin 2 a 1 
f= (1 -e*) ' a' 1 

and substituting the value of a already found [163], 

r , 2X v 1 sin 2 a . 9X 

(*— -0 = /v * (*fr-v\ 

and in like manner for the hyperbola 

, „ , v 1 sin 2 a , . _ . 

y~i)= v frA-2/r). 

Now, the first members of these equations being positive, 
the second members must be so also, which gives for the 
ellipse, etc., ^ 2 < 2/r, and for the hyperbola v* > 2fr, as be- 
fore ; the value of — tt~* — , which contains the variable angle, 
f r 

being always positive, whilst the factor in parenthesis deter- 
mines the sign, and therefore the species of the curve. Hence 



DYNAMICS. 189 

the value of a does not affect the species of the orbit, as already 
seen. 

[165.] If we assume, as the unit velocity, in the case of a 
body launched into space to assume an orbit about the sun, 
the velocity necessary to maintain a circle, at the distance r, 
we shall then have, as the conditions for the different orbits 
for the same distance, 



the circle, 


v—\\ 


the parabola, 


v = V2; 


the ellipse, 


v < V2; 


the hyperbola, 


v > V2". 



It will be observed, moreover, that for the circle and parab- 
ola definite velocities are required, whilst the condition for 
the hyperbola is simply that the velocity, under our assump- 
tion of the unit velocity, shall be greater than V2, and the 
velocity for the ellipse ranges between o and V2, including 
the velocity of the circle, as it should do. The chances would 
therefore be against the 'circular or parabolic orbits, on the 
hypothesis that the bodies should be projected at random. 

[166.] To find an expression for the velocity in the various 
orbits, in terms of the central force, radius-vector, and major axis 
of the curves. 

From section [163] we have immediately, 

for the ellipse, etc., / = 



2 a r 



whence ^ 2 = 2/r(i ); 

\ 2 a' 

and for the hyperbola, /: 



2ar4-r* 



whence v* = 2/r f 1 -| J 



i90 ELEMENTS OF MECHANICS. 

It will not be difficult for the student to see that the con- 
clusions of section [163] could have been deduced more 
directly from these equations. For taking the equation 



v * = 2fr ( i - Q' 



applicable to the first three conic sections, it is plain that for 

f T T 

the circle — = -J-, for the parabola — = 0, and for the ellipse — 
2 a 2 a 2 a 

equals some proper fraction, and positive. Making these sub- 
stitutions, we should plainly have, 

For the circle ; v 2 —fr ; 

For the parabola, v 2 = 2fr ; 

For the ellipse, v 2 < 2/r ; 

and from the second equation, in like manner, 

For the hyperbola, v 2 > 2/r. 

[167.] To find a7i expression for the semi-conjugate axis. 
We have, section [164], for the ellipse the formula 



(i-O a' 

, , 2X v 2 sin 2 a 

whence (1 — e) — . 

fa 

72 

But (1 — e 2 ) = — -, which gives 

b 2 v 2 sin 2 a .„ 

or b 2 — 



f" f 



Or if we denote by k the acceleration of the central force 
upon the unit mass at the unit distance, we have 



r I 



DYNAMICS. I9I 

k a 

or f— -s and £ 2 = r 2 ^ 2 sin 2 aX 7 , 

r k 

and b = rv sin « y -. 

Whence it appears that the minor axis, unlike the major, does de- 
pend for its value upon the angle of projection. The same value 
for b results also for the hyperbola. 

[168.] Let us now suppose a planet to be projected, with 
a velocity insufficient to maintain a circle, and at right angles 
to the radius-vector r, joining it with the sun. It would then 
move in an elliptic orbit, the initial point of the motion being 
the aphelion. If the velocity should be decreased until it 
approached very near to zero, the orbit would become very 
eccentric, as would be shown also by the equation 

b = r v sin a \ T = r v V -r ; 
k k 

and the perihelion would be very near the centre of force, 
the body being deflected by the central force, under the law 
of the orbit, into a sharp curve, at its perihelion. 

But if v = o, the body would fall directly toward the sun, 
which we may suppose to be replaced by a mere centre of 
attraction; and passing beyond it, no deflection into a curve 
being possible, because the lines of direction of the force and 
motion are coincident, it would rise to an equal distance on 
the opposite side, and continue to vibrate backward and for- 
ward in the same straight line. The length of this line would 
be equal to the circumference of a very eccentric elliptic 
orbit ; and is in truth but the development of such an orbit. 
Regarding, as in section [165], the velocity necessary to main- 
tain a circle as unity, and supposing now the initial velocity 
to begin to increase, the perihelion would begin to move off 
from the focus in which was situated the central force, the 
elliptic orbits becoming less and less eccentric, until, when 



192 ELEMENTS OF MECHANICS. 

v = 1, the orbit would become a circle, with the aphelion and 
perihelion points at equal distances from the central force, 
now in the centre of the circle. If we continue to increase 
the velocity of projection beyond unity, the circular would 
pass into an elliptic orbit, becoming- more and more eccentric 
as v increases, the initial aphelion point becoming an initial 
perihelion point. But when v = V2, the ellipse would fall 
into the parabola with the same vertex and focus, a single 
curve, and immediately, upon the velocity becoming greater 
than 4/2, the hyperbolic orbit would begin. 

We see from this also that the velocity requisite for an 
elliptic orbit will be, 

When the initial point is at aplielion, v > o and < 1 ; 
When the initial point is at perihelion, v > 1 and < V2. 

[169.] If T denote the time of revolution in any orbit A B (Fig. 
95), whose mean distance is a, the time required for the revolving- 
body to fall into the centre of force, under the law of the force ; 

T 

through the mean distance, zvill be equal to — -=• 

For suppose the body placed at P to receive the very 
slightest impulse at right angles to Pf By the preceding- 
section, it would move in a very eccentric orbit PQ, with the 
focus f indefinitely near the extremity of the major axis ; and 

the mean distance in this orbit would be -• If then T. denote 

2 

the time in the orbit P Q, Kepler s Third Law will give 

77 : T- :: ~ : a" :: \ : 1, and T x = ^=. 
8 8 ' 4/8 

But the time of the fall from P to /, which we may denote by 
x, is obviously equal to half the time of revolution in the in- 

definitely eccentric orbit PQ ; that is, equal to — • 



DYNAMICS. 193 

We have therefore 



x 



VS 4 V2 
as was to be shown. 

[170.] To find an expression for the angular velocity, or ve- 
locity of a point situated on the radius-vector, at the unit -distance 
from the centre of force. 

If we denote the angular velocity by 6, it will be obvious, 
by simply drawing a diagram, that in the unit of time 

r 6 = v sin a. 

From a preceding article, we have 

b = rv sin a y —, 

7 2 a i/a j b Vk 1 

or b = r 2 6 y - and = — — x — , 

k tf a r 

from which it is seen that the angular velocity varies inversely 
as the square of the radius-vector. 

This equation may, however, be put under another form 
thus : from the equation 

J a(i—e 2 )r J a(e — \)r 

we have, by substituting -- for (1 — e 2 ) or (V 2 — 1), 

But/r 2 = k, section [167]. Hence 



*£? A A^bA and b\/ k - 

A 



k = *-»-, 4A*=b*- and bV- = 2A, 
b ^ a a 



and, by substitution, 6 = 

j r 



194 ELEMENTS OF MECHANICS. 

It may be well also to call attention to the simple equation 
for the area described in the unit of time, or 

2 a 
See latter portion of [160]. 

[171.] To find an expression for the velocity at any point in 
the orbit, in terms of the central force at the unit distance, and 
a and r. 

We have, section [166], 

and since k = /r a , we have 

S k A i 2k ( T \ 

fr = ~ and v= — i— — , 
J r r \ 2a) 

as required. 

But, section [164], we have for the measure of the central 

force at the distance r the equation 

/- ^ 



(1 -e 2 ) rd 
Now if r = 1, then 

f — k and £ - 



«(i-0« 



the value of the central force at the unit distance upon the unit 
mass. Hence, by substitution, 



v = 



I r \ _ 4 A* 12 a — r\ 

V~2~al~~ a' (1 - S) \ ~) 



a(i—e*)r\ 2 a) a* (1 — e 1 ) 
and similarly, for the hyperbola, 

4 A' 2 /2a-\-r" 



V*: 



d(e*-i) 



/-£ 


£ ^ 2 

or - = -L- ; 


<-i_ 


4-4" 


'«(l —f)r 



DYNAMICS. 195 

In the revolution of a body in a circular orbit, section 
[151], with a radius r, we have 



whence 



By comparing the velocity in the elliptic orbit for the 
same distance r as found above, we have 

v : v, :: : - :: (2 a — r) : a; 

ar r 



that is to say, the square of the velocity in the ellipse, is to the 
square of the velocity in the circle at the same distance, as the dis- 
tance of the revolving body from the empty focus of the ellipse, is 
to the semi-major axis. It is obvious that at the extremity of 
the minor axis 2 a — r = a, and the velocities in the two orbits 
would be equal, as has already been shown [157]. 

[172.] Of the elliptic orbit when the locus of the force is the 
centre of the ellipse. 

Resuming the general equation for the central force, sec- 
tion [153], viz., . 



F = Mv*p 2 -^- z , or f=v*f 



Rp« J r Rp 3 ' 

we have in this case (Appendix A), 

v'p'r 



Rp 3 = a 2 b 2 and /= 



~2 7,2 ' 

a 



Hence, since v 2 p i is constant, we conclude that, when a body 
revolves in an elliptic orbit, with the central force located at the 
centre of the ellipse, this force varies directly as the radius-vector, 
or distance of the body. 



I96 ELEMENTS OF MECHANICS. 

[173.] To find the periodic time T in the orbit. 

Since the central force varies directly as the distance, if k 
denote the acceleration of that force at the unit distance, we 
have 

/: k :: r : 1, or f=kr. 

Hence, by substitution in the equation of the preceding sec- 
tion, 

, v 2 f , abVh 

&=-— and v == , 

a p 

for the velocity at any point in the orbit. If we suppose this 
point to be the extremity of the minor axis, then p = b and 
v — a Vk. But if A denote the area described by the body 
from this point in the unit of time, we have 

. _ v p v b 

~ 2 ~~ 2 ' 

and as neretofore we shall have, for the whole time in the 
ellipse, 

nab 27tab 2 n a 2 7t a 2 7t 

~ A vb v ~ aVk ~" Vk 

It appears, therefore, that, when the centre of force is at the 
centre of the ellipse, the periodic time is constant, and inversely as 
the square root of the force, at the unit distance, and consequently 
the time zvill be the same for all orbits around the same central 
force. 

[174.] In the foregoing discussion, the use of the calculus 
has been intentionally avoided, with the view of bringing this 
interesting subject within the reach of those who have only 
mastered the elements of mechanics and analytical geometry. 
A single example is, however, given here, to serve as an illus- 
tration simply, for those who may desire to test the law of 
the force that would obtain for different orbital curves. 

Let it be required to determine the law of the force which 
would cause a body receiving an initial impulse, to revolve in the 
logarithmic spiral. 



DYNAMICS. 1 97 

Having the general formula for all curves [153], viz., 

it is required to determine Rp z in terms of r. The formula 
found in almost any work on the calculus, for the perpen- 
dicular/ upon the tangent, the curve being referred to polar 
co-ordinates, is 

P 



and for the radius of curvature, 

dr 
R = r d}' 

Assuming the equation of the logarithmic spiral 

r = a e , 
we have log r = 6 log a ; 

and differentiating with regard to 0, 

r* 
Hence p = , 

Vr' + rnog'a 

and differentiating with regard to p, we have 
2p(i+\og 2 a) = 2r-^ J 



SS=^+^«> 



<p 

Hence R=r—=p(i-\- log 2 a) 

dp 

and Rp*=p l [i + log 5 a) ; 



ie/ '/(i+iog**) 



I98 ELEMENTS OF MECHANICS. 

But * = (i+iog^r 

Hence /= v*?(i + log 2 *) . ± 

whence it is seen that ^ central force varies inversely as the 
cube of the distance. 

The student is now prepared to solve other examples by 
the same method, as, for example, the lemniscata of Bernoulli, 
in which, as an orbit, the force would vary inversely as the seventh 
power of the distance ; and by assuming the general polar 
equation of the conic sections, the results already obtained in 
section [155] could be confirmed. 



Of Centrifugal Forces. 

[I75-] In section [151, Cor. 3] it was shown that the cen- 
trifugal force of a mass revolving with uniform velocity in a 
circular orbit was represented by the equation 

Wv> 
F— , 

in which the whole mass was supposed to be concentrated in 
one point. 

It is proposed here to treat of the centrifugal force of a 
pendulum vibrating in its arc, the centrifugal force at differ- 
ent latitudes upon the earth's surface, and of the centrifugal 
force of extended masses. 

[176.] A simple pendulum of weight W vibrates in the arc 
Abe (Fig. 96). It is required to find an expression for the ten- 
sion of the cord at any point m in the arc of descent from c to b. 

The velocity at m equal to that accumulated by a heavy 
body falling freely through nd, section [128, Cor.], will be 
equal to 



V2g(bn — bd) — V2g\(r — rcosa)- (r— rcos(a — ar T )J, 



or v — V 2g J r cos (a — or x ) — r cos oc\ 



DYNAMICS. 



I 99 



This value for the velocity, being substituted in the for- 
mula for the centrifugal force [175], 
gives 

F= 2 Wjcos {a — a x ) — cos a\. 

But this being the centrifugal force 

simply, and the same as if the body 

revolved in a horizontal circle, the 

tension due to the weight itself is still 

to be added. Resolving the weight 

at the point m, as shown in the diagram, into the components 

mt — x and my tangential, the value of m t or x will be the 

additional tension due to the weight alone. Similar triangles 

s m t and m do give, since the angle s m t = a — a v 




and hence 



x = W cos (a — or,), 
F = 3 J^cos (a — «,)— 2 Wcos a. 



(a) Let the arc of vibration be 180 , or a = 90 , and sup- 
pose also ol x = 90 ; the formula will then give the tension at 
the lowest point b of the arc. But under this supposition we 
have 

cos (a — tfj) = cos 0=1 and cos a == cos 90 = o, 

and F= 3 W. 

(b) Let a = 90 , and a x = 30 . Then 

cos (a — a t ) = cos 6o° = sin 30 = i, and F= f W, 

and the tension at 6o° from the lowest point is equal to J the 
tension at that point. 

(c) Let it be required to find the point at which the ten- 
sion is just equal to the weight W, when a = 90 . In this 
case we have 

F= W= $W cos (a — or,), 

or cos (a — a l )=i= .3333, etc. 



200 ELEMENTS OF MECHANICS. 

Referring to a table of natural cosines, we find 
a — ar, =7o°32 / 15", 

the distance from the lowest point b to the point at which 
the tension is equal to the weight. 

It is obvious that if Ave assume a and a x of any arbitrary 
values, the formula, by means of the table of natural cosines, 
can be applied generally. 

Example. — A boy weighing 100 lbs. rises into a horizontal 
position upon a swing. Required the tension of the ropes 
when he is at the lowest point of the circle. 



OF THE EFFECTS OF CENTRIFUGAL FORCE UPON BODIES ON 
THE EARTH'S SURFACE. 

[177.] The centrifugal force at any point on the earth's sur- 
face, is equal to the centrifugal force at the equator, mutiplied by 
the cosine of the latitude of the place. 

Let N S (Fig. 97) be the axis of the earth, E Q the equator, 
and B D parallel to E Q, a parallel 
of latitude, the earth being re- 
garded as a sphere. During one ^ 
revolution on its axis a body 
placed at E would describe the 
equatorial circle, with a radius 
equal to E C '= R, and a body at 
B y a parallel of latitude, with a 
radius equal to B C x = r. 

Hence if F denote the centrifu- 
gal force at the equator, and i 7 , that 
at the point B, we shall have, sec- 
tion [151, Cor. 2], 

R 




Fig. 97. 






r 

Tr 



DYNAMICS. 20 1 

or, since the time for describing the orbits is the same for 
both bodies, 

F\F X \\ R\r. 

Let a represent the latitude of the place, and there results 

r = R cos a. 

Hence F : F x : : R : R cos a : : 1 : cos a, 

or F x = .Fcos oc, 

as enunciated. 

If the body be placed at the pole, 

a = 90 , cos a = o, and F x = o. 

If at 30 from the pole, 

oc — 6o°, cos a = \, and F X — \F. 

[178.] The diminution of the force of gravity at any place on 
the earth's surface, is equal to the force of gravity at the equator, 
multiplied by the square of the cosine of the latitude of the place. 

Although the effect of the centrifugal force is to diminish 
the force of gravity at all points of the earth's surface except 
at the poles, yet it is only at the equator that these two forces 
act in direct opposition to each other. At the point B (Fig. 
97) let Bf represent the direction and intensity of the cen- 
trifugal force. Resolving this force into Bh tangent and B g 
perpendicular to the earth's surface at that point, the latter 
force alone acts in opposition to the force of gravity, which 
is therefore diminished by the value of Bg. But 

B g— Bf cos a; 

and since, by the preceding article, 

Bf — F 1 = Fcos a, 

we have Bg = Fcos* a, 

which was to be shown. 



202 ELEMENTS OF MECHANICS. 

[179.] At the equator the centrifugal force is equal to -g-J-g- part 
of the absolute value of the force of gravity. 

Let the intensity of the force of gravity at the equator, as 
determined by the method explained in section [144], be de- 
noted by g, the absolute intensity of the same force, on the 
hypothesis that the earth does not induce a centrifugal force 
by rotation on its axis, by G, and the centrifugal force by F. 

We shall then obviously have 

G=g+F. 

The value of g found as indicated is 32.0861 feet, and the 
value of F may be determined, section [151], by the general 
equation 

r - T 2 ' 

In the present case r represents the equatorial radius of the 
earth, equal to 20,920,300 feet, and T the time of the earth's 
rotation on its axis, equal to 86.164 seconds of mean solar time. 
Substituting these values of T and r, we find, after reduction, 

F — o. 1 1 1 2 feet, 
and hence 

G =g+ F— 32.0861 +0.1 1 12 = 32.1973 feet, 

.F0.1112 1 1 

or — = — — = — - and F = ■=-- G, 

G 32.1973 289 289 

as stated. 

It will be observed that G here denotes the acceleration due 
to the force of gravity, or the force of gravity itself, if this ac- 
celeration be assumed as its measure, in relation to other like 
forces ; and since the diminution of the absolute force of grav- 
ity in any latitude will be equal to 

Fcos 2 a = 0.1 1 12 cos 2 ot, 

in which a [177] denotes the latitude, a similar method to that 
just given will make known to us the force of gravity for any 
latitude, on the hypothesis that the earth does not revolve on 
its axis. 



D YNAMICS. 203 

Experiments go to show that the intensity of gravity at 
the poles would exceed that at the equator by the T ^ part of 
G. Of this excess, as we have already seen, the ^irw P ar * °^ 
G is due to the action of the centrifugal force at the equator, 
and hence what remains is due to the spheroidal form oi the 
earth. 

[180.] Let it be required to find the time in whicJi it would be 
necessary for the earth to revolve \ in order that the force of grav- 
ity at the equator would be just equilibrated by the centrifugal 
force. 

Recurring to section [151, Cor 2, c\ we have for the cen- 
trifugal forces of the same body revolving in different orbits 
the proportion 

c c r r x 

F F " • - 

If the orbits become the same, or r — r 1} we have 

F:F X :: T, 2 : T\ 

Let the orbit be the equatorial circumference of the earth, and 
T the actual time of the earth's rotation. The force F will 
then be the centrifugal force corresponding to the time T; 
and if T y be the required time of revolution when the cen- 
trifugal force F 1 corresponding shall become equal to G t the 
force of gravity, the last proportion will give 



F : G :: T^ : T\ or T, = V ^ ■ T= V -~ • T= - 



G 



T 



/ 



that is to say, the required time will be equal to -^ of the 
actual time of 24 hours. 

OF THE CENTRIFUGAL FORCES OF EXTENDED MASSES. 

[181.] In the preceding investigations, we have supposed 
the mass in which the centrifugal force was generated to be 
simply a material particle, and all portions of it equally dis- 
tant from the axis or centre of motion. 



204 ELEMENTS 0E MECHANICS. 

But in an extended mass supposed to revolve about an 
axis, the different elements of the body, being at unequal dis- 
tances from the axis, will generate different degrees of cen- 
trifugal force, and the centrifugal force of the whole mass 
will be equal to the resultant of the centrifugal forces of all 
of its elements. 

It is proposed to show that the centrifugal force of the body 
will be the same as if its whole mass were concentrated at its cen- 
tre of gravity. 

Let B (Fig. 98) be a mass which is supposed to revolve 
about an axis A Q, and all the elementary particles of which 
are conceived to move in planes perpendicular to this axis. 
Conceive the whole mass to be divided 
into an indefinite number of infinitely 
small linear elements parallel to the axis 
A Q, and suppose a plane Y A X to pass 
through its centre of gravity, perpendic- 
ular to and cutting the axis A Q at the 
point A, making a section 55 x of the \ 
body. Since each linear element of the 9 

mass is parallel to A Q, and equidistant FlG * 98, 

from it at all points, the centrifugal force generated by any 
one of them will obviously be the same as if the whole mass of 
the element were concentrated at any point in its own length 
or line of direction. 

We may therefore suppose the mass of each element to be 
at the point in which it pierces the section 5 S t made by the 
secant plane perpendicular to A Q. 

Let m denote the mass of the element piercing the section 
at the point m. Denote its distance from A by r, and its co- 
ordinates, referred to A X and A Y as rectangular axes, by x 
and y. 

The centrifugal force of the elementary mass m, whose ve- 
locity is v, will then be, section [151, Cor. 3], 



Y 

I 








Y 



DYNAMICS. 205 

Denoting by V x the angular velocity of the body, or the 
velocity of a particle at the unit's distance from the axis, we shall 
have the linear velocity of m equal to 

v = r V x . 

Whence, by substitution, the centrifugal force of m in terms 
of its angular velocity will be 

m r V 2 . 



Representing this force by m n, in the diagram, and resolving 
it into its two components m and mp y parallel to the axes of 
X and Y, we shall have for the components respectively par- 
allel to these axes the expressions 

m r V x cos mAX and m r V 2 sin m A X. 

But referring to the diagram, 

cosmAX=- and sinmA X = ^-. 
r r 

These components therefore become 

mx V x and my V x . 

Since similar expressions for the centrifugal force of each ele- 
ment may in like manner be obtained, we have for the result- 
ants of these forces parallel to the axes, which may be denoted 
by X and Y, the expressions 

X = 2 m x V 2 and V = 2 my V 2 , 
or X= V 2 2mx and Y= V 2 2 my. 

If we denote by x x and y 1 the co-ordinates of the centre ot 
gravity of the whole mass M, and by r x its distance from A, 
we have 

2 m x = Mx x and 2 my = My x . 
Whence X = M V 2 x x and Y = M V? y x . 



206 ELEMENTS OF MECHANICS. 

Representing by V the velocity of the centre of gravity of 
the body, we have 

F 3 



r=r x v x , v: = 



r, 



and X = — - 1 and Y— j-A 

r* r x 

The resultant of the forces X and Y, which are the com- 
ponents of the centrifugal forces, in the direction of the two 
axes, at right angles to each other, will plainly be the result- 
ant centrifugal force of the whole system of elements, or of 
the whole body. Denoting this resultant by R, we have 



R = VX* + F 2 = -^-V Vx? +//. 
r \ 

But (x; + y*) = r;\ 

M F 2 
Hence R — . 

But this, section [151, Cor. 3], is the centrifugal force of the 
whole mass M collected at its centre of gravity. Hence the 
truth of the proposition as stated. 

EXAMPLES. 

(a) Two spherical bodies weighing respectively 3 and 9 
lbs. are fixed to the extremities of an inflexible line. Required 
the point in the line about which, as a pivot, if the two masses 
revolve, the centrifugal forces will be equal, 

Denoting the lengths of the respective arms from the re- 
quired point to the centres of the respective spheres by r and 
r v and the velocities of these centres by v and v lt the centrifu- 
gal forces will be given by the expressions 

3^ and ^_ 



D YNAMICS. 207 

Equating these expressions, we have 

3 v* 9 .v* v[__ $v?_ 



^ r gr 



But the revolutions being in the same time, the velocities will 
be proportional to the radii, or we shall have 



v : v 1 : : r : r v and v* = 
whence, by substitution, 



v* r? 



v* 3 v* r x 



or r = 3 r x , 



and the arms will be to each other as 3 to 1, or inversely as 
the weights, 3 to 9. 77/<? required poi?it is therefore coincident 
with the centre of gravity of the two masses. The same will ob- 
viously be true for any two masses rigidly connected by their 
centres of gravity. 

(&) An equilateral triangle, cut from a metallic plate, whose 
sides are equal to one foot, and the weight to two pounds, re- 
volves at the end of a cord attached to one vertex, 3 feet in 
length, the centre of gravity moving with a velocity of 20 
feet per second. 

Required the tension of the cord in pounds. 

(c) A semi-ellipse cut from a metallic plate, with the semi- 
axes equal respectively to 3 feet and 2 feet, revolves about the 
conjugate axis. Supposing the plate to weigh 1 pound to 
the square foot, it is required to find the number of revolu- 
tions per second, when the centrifugal force is just equal to 
the weight. 

The conjugate axis is here considered the centre line of 
the straight edge of the plate. 



208 



ELEMENTS OF MECHANICS. 



THE CONICAL PENDULUM. 

[182.] The conical pendulum (Fig. 99) consists of a weight 
or ponderable point attached to the extremity of a very fine 
cord, and made so to vibrate that the cord will generate the 
convex surface of a right cone, whilst the weight revolves in 
a horizontal circle. 

Referring to the diagram, it is evident that the body B, re- 
volving in the horizontal circle B A D y will be, as it were, 
solicited by three forces, viz., its 
own weight, represented by B G, 
the centrifugal force at the point B, 
represented by B F, and the conse- 
quent tension of the cord, equal to 
their resultant B R, in the direction 
B V. Let these forces be repre- 
sented respectively by G, F, and R. 

Since the centrifugal force F 
would be equilibrated by an equal 
opposite force in the direction B C, 
we may conceive the two forces G 
and R to be equivalent in effect to a central force at C, equal 
to F. Moreover, any change in the velocity of revolution in 
giving a new value to the centrifugal force F would also give 
a new value and direction to the resultant R ; and since the sus- 
pending cord V B must preserve the same direction as this 
resultant, the weight B must of necessity rise to a higher or 
descend to a lower orbit as the centrifugal force is increased 
or diminished by a change in the orbital velocity of the body. 

If, instead of supposing the body to revolve in its orbit by 
means of the cord VB, we conceive a second conical surface, 
whose section is V^BD, and which is at right angles to the 
elements of the first surface, the two surfaces would intersect 
in the same circle BAD, and we might without change of 
effect conceive the body to move in the same orbit, supported 
by the inner surface of the lower cone, the reaction of which 
surface, in a direction normal to itself, taking the place of the 
tension of the cord, in the case of the suspension of the body. 




Fig. 99. 



DYNAMICS. 209 

[183.] To find an expression for the time of vibration of the 
eonieal pendulum. 

Denoting the velocity of revolution by V, the weight of 
the body by G, the central or centrifugal force by F f the ra- 
dius of the orbit B C by r, and the altitude V C of the cone by 
h, we shall have> section [151, Cor. 3], 

V 2 V 2 

F : G : : — : g or — : g : : B F ' : B G : : r : h. 



r 



r *l n „A Tr_„\/g 



Whence V 2 — -7- and V= r r 

h h 

But the time of revolution is equal to 

2 n r A /~h 

t = -jrr =2 ay--. 

v g 

This expression, which involves the altitude of the cone 
only, and constants, shows that the time of vibration is the 
same for all cones of the same altitude h, whatever may be 
the length of the cord of suspension. It should be observed, 

also, that the time of revolution 2 n V - is just equal to twice 

the time of vibration of a pendulum equal in length to the 
altitude of the cone, the arc of vibration [139] being assumed 
to be small. 

If we conceive the velocity of revolution in the formula 

V= r\ %- to increase, it is obvious that r will increase and 
h 

the altitude of the cone h decrease ; and finally, when h = o, 

or when the body is supposed to revolve in a horizontal circle, 

the expression for the required velocity becomes 

V=rV ^ = oo, 
o 

and that for the time of revolution 

t = 27tV - = 0, 
g 

as might have been anticipated. 



210 ELEMENTS OF MECHANICS. 

Under this supposition, also, the centrifugal force would 

be [150] 

f— — =00. 

That is to say, the centrifugal force requisite to raise the 
weight to a level with the point of suspension is infinite in 
value. On similar principles it follows that to stretch any 
ponderable line into a horizontal position that would be accu- 
rately straight would require an infinite force. 

[184.] It has been assumed in the above discussion that 
the revolving body is kept in motion in a horizontal circle ; 
but even if such a motion could be imparted to it for one 
revolution, it would, if left free to assume its orbit under the 
action of the three forces by which it is influenced, practically 
tend to an elliptical motion, though the orbit would no longer 
lie in one plane, since the extremities of the transverse axis 
would be more elevated than those of the conjugate. 

If we substitute the value of V 2 in the formula for the cen- 
trifugal or central force, we have 

J r ~ rh~ h y 

from which it is seen that the central force varies directly as 
the radius r, and inversely as the altitude of the cone h. But, as 
was shown in section [172], when a body moves in one plane, 
subject to the action of a central force which varies directly 
as the distance simply, its orbit will be an ellipse. Hence in 
the present case, since one element of the law is the same, 
the tendency to an elliptical motion is explained. When the 
orbit is small, it is obvious that a change in the value of r 
produces relatively a small change in the value of //, and the 
second element in the law of the central force has little effect, 
causing a nearer approximation to an elliptical orbit. 

[185.] Let it be required to determine the i?tcli?iation of a 
curved roadway, when the radius of curvature and the velocity of 



D YNAMICS. 2 1 1 

the vehicle is given, so that the pressure of the moving body may 
be normal to the surface. 

Let BAD (Fig. 99) be the curvature of the roadway. The 
proper inclination of the track to the horizon for the velocity 
V would then be equal to the angle C B V it since the pressure 
of the revolving weight, whose velocity is V, is normal to B V x . 

But CB V x = 90 -VBC=C VB. 

From the last article we find 

n — yi 

But h : r :: cos C VB : sin C VB, 

cos C VB ^ Tr „ 

and h = r —. — „ ir n = r cot C VB. 

sin C VB 

Equating these two values of h, we have 

?j£ =r cot CVB, or cot C VB = ^, 

whence C VB becomes known. 

Example. — A vehicle supposed to be moving at the rate 
of 20 miles per hour, or 29.33 feet per second, passes the curve 
of a roadway whose radius of curvature is equal to 600 feet. 
Required the cross inclination of the track to the horizon, in 
order that the pressure of the vehicle may be normal to its 
surface. 

In this example we have 

r — 600 ft., g — 32^ ft. per second, 
and V= 29.33 ft. per second. 

Hence cot C VB = r £ = -'|^° = 22.43. 

V 2 860.25 ^ J 



212 ELEMENTS OF MECHANICS. 

Introducing the radius R of the ordinary table of logarithms, 
observing that in deriving the formula radius has been con- 
sidered as unity, we have 

cot C VB — R X 22.43, 

and, passing to logarithms, 

log cot C VB = log R -f- log 22.43 = 10 + 1.350829, 

or log cot CVB = 1 1.350829, and CVB=2° 33' 10'. 

Note. — The above solution would not accurately apply for 
determining the proper elevation of the outer rail in passing 
a curve in a railroad-track. In this case, since the wheels 
move upon an axis to which they are firmly attached, and the 
distance to be passed over is greater on the outer than on the 
inner rail of the curve, it becomes necessary to give to the 
wheels a particular form in order to bring up the outer wheel 
in proper time, and to prevent its dragging or tending to run 
the train off the track in a rectilinear direction. To obviate 
this difficulty the wheels are made of a slightly conical form, 
having a greater diameter near the flange, on the inside of 
the rails, than toward the outer rim. Hence in passing a 
curve, the tendency of the train being to pursue a rectilinear 
course, the flange of the outer wheel will be brought closer 
to the rail, thereby slightly increasing the diameter, whilst at 
the same time the flange of the inner wheel tends to separate 
from the rail, and somewhat decrease its virtual diameter. 
The result is that the excess of the circumference of the outer 
wheel, which comes in contact with the rail, not only gives 
the train a curvilinear motion, but compensates for the greater 
length of arc on the outer rail. 

Since the elevation of the outer and depression of the 
inner wheel places the weight of the train upon an inclined 
plane corresponding to that which would be produced by the 
elevation of the outer rail, the necessity for this elevation is 
in some degree obviated, and hence in the example above, 
applied to a railroad curve, the angle C VB would be some- 



DYNAMICS. 



213 



what less than the value found above, depending upon the 
form of the wheel, the amount of lateral displacement of the 
train, and in a slight degree upon the width of the track. But 
this subject belongs more appropriately to works on civil en- 
gineering, and for the case under consideration the student is 
referred to Mahan's " Civil Engineering," Note IX., Supple- 
ment. 



[186.] Required the form of a curved surface such that a 
body would revolve in a given periodic time in any horizon- 
tal circle cut from the surface without a tendency either to 
rise or fall. 

Let D B C (Fig. 100) be a section of the required surface, 
whose axis is AB; and PQ and 
P\ Qi two horizontal circles de- 
scribed in the same time /. Draw 
A P and E P v respectively perpen- 
dicular to the curve at the points 
Pand P v Then will A O = h and 
E 1 = h t be the altitudes of the 
cones of revolution whose ele- 
ments are perpendicular to the re- 
quired surface. 

Now if / be the common time 
of revolution, we shall have for its value, as shown in section 
[183], the expressions 

if 1 

g 




g 



Whence 



h = Ju 



which is therefore the condition to be fulfilled in order that 
the times of revolution in the two orbits shall be equal. Since 
the same could be shown to be necessary for the altitudes of 
all other cones whose bases are the horizontal circles of revo- 
lution, the equation h — h x becomes a general one, necessary 
to fulfil the required conditions of the problem. 



214 ELEMENTS OF MECHANICS. 

But h and //, are subnormals to the curve. Hence the 
curve section must be such that its subnormals are constant, 
which is true only of the parabola. The required surface 
must therefore be that of a paraboloid of revolution. 



Moment of Inertia. 



[187.] If we suppose a system of ponderable particles 
m, m v m v etc. (Fig. 101), rigidly connected by the line C B, to 
be made to rotate in one plane about an axis perpendicular to 
this plane through C, by a force F applied at £, the velocities 
of the various particles would all be different, and depend 
upon their respective distances from the centre of motion C. 

Each particle would also offer a resistance to motion due 
to its inertia, which would be di- 
rectly proportional to its actual ve- 
locity. Whatever this resistance 







« *t Jit 


2* 11 


Fig. ioi. 





might be, if it were in each case 
multiplied by the distance of the 
given particle from C, the result 
would be the moment of resistance, or moment of inertia, of the 
particle with respect to the point C, as opposed to the mo- 
ment of the force F in the opposite direction. 

Let a be a particle at the unit's distance from C, and let the 
arc a denote its velocity, being so many degrees or parts of 
a degree of the circumference, whose radius is unity, passed 
over in a unit of time, and called the angular velocity. 

If now we suppose the particles m, m v m v etc., to be at 
distances r, r lf r 3 , etc., from C, their actual velocities will be 
r a, r x a, r 2 a y etc. ; and since the resistance to motion of each 
particle, or the inertia developed, would be proportional to 
its mass multiplied by the velocity accumulated, we might 
denote these resistances by the expressions 

m r a, m 1 r x a, m^ r 2 a, etc., 



DYNAMICS. ' 215 

and their moments with respect to C as a centre of moments 
would be 

m r 1 a, m x r* a, m % r* a, etc. 

If we assume, however, as the unit of measure for mo- 
ments the moment of resistance of the unit particle, placed at 
a unit's distance from the axis of motion and moving with a 
unit velocity, then a = 1, and we have for the sum of the 
moments of inertia, or resistance, of all the particles 

m r* -f- «, r* -f- w„ ?' 2 2 + etc. = 2 m r 2 

for the moment of inertia of the whole system, which is the 
moment of resistance to rotary motion in terms of the unit 
adopted. 

It should be observed here that in rectilinear motion it has 
been assumed that the inertia developed in a mass is equal to 
the product of the mass by the velocity, because the inertia 
developed in the unit mass in receiving the unit velocity is 
adopted as the unit of measure for inertia. In like manner in 
rotary motion the unit mass, receiving the unit velocity, de- 
velops a resistance assumed as a unit of measure, and the 
inertia developed in a particle subject to rotary motion, mul- 
tiplied by the distance of the particle from the axis of mo- 
ments, gives the moment of inertia of the particle. 

Again, if we conceive the whole mass of the particles equal 
2 m = M to be concentrated at B, and whilst free to move in 
any direction to receive the impulse F, imparting to it some 
velocity Fin a unit of time, then M V would be the measure 
of the force F\ and if we denote the distance C B by R, the 
moment of F about C would be 

MVR. 

The angular velocity being supposed to be uniform, it is 
obvious that the moment of F f or M VR, and the sum of the 
resisting moments of the particles moving with this angular 
velocity must be in dynamic equilibrium, else the velocity 
would become either greater or less until this equilibrium was 



2l6 ELEMENTS OF MECHANICS. 

brought about; in which case it is plain we should have the 
sum of the moments of the parts equal to the moment of F, or 

{in r* + m x r? + m % r* + etc.) a = 2 m r\ a = M VR, 

MVR 

and a = •= — 

2, m r 

an expression for the angular velocity. 

In rectilinear motion the expression M V denotes the mo- 
mentum or quantity of motion of a mass M moving with a 
velocity V [4]. 

The inertia of the mass being proportional to the mass 

itself, or equal to the inertia developed in receiving the unit 

velocity, would be represented by M X 1, or M. In this case 

MV 
the expression V= — 77- shows that the velocity of motion is 

equal to the quantity of motion divided by the inertia or re- 
sistance to motion for the unit velocity. 
In like manner, in rotary motion, we have 

MVR 

in which expresssion the numerator represents the quantity 
of motion impressed upon the system, and the ^mr 1 the re- 
ristance to rotary motion due to the whole mass for the unit 
angular velocity. 

[188.] In treating of the rotation of a solid body it be- 
comes necessary, therefore, to take account of that portion of 
the moving force which is consumed in overcoming the in- 
ertia of the system of particles of which the body is composed, 
and which cannot therefore conduce to practical results in 
overcoming resistances. 

In order to determine the value of the expression ^mr^, 
the resistance due to inertia, for different bodies, rotating upon 
an axis, where the subject has been developed to any extent, 
the methods of the calculus are usually adopted ; but, in con- 



DYNAMICS. 



217 




Fig. 102. 



formity with the plan of the present treatise, all such examples 
will be solved by algebraic methods. 

[189.] If the moment of inertia of a mass with respect to an 
axis passing through its centre of gravity be given, its moment of 
inertia with respect to any other parallel axis may also be deter- 
mined. 

Let A B (Fig. 102) be a section of the given mass, produced 
by the plane of XZ passing through C, its centre of gravity, 
supposed to be situated at the ori- 
gin of three rectangular co-ordinate 
axes. Also let Yn be an axis coin- 
ciding with the axis of Y, and hence 
perpendicular to the plane of the 
section A B. Let D E, passing 
through the point E of the section 
at a distance from C equal to a, be 
a parallel axis. Taking now any 
element of the plane A B, as m, de- 
noting its distance from E and C respectively by r and r v and 
its abscissa C x by x, we have, from the triangle E Cm, 

r' = r," -)- a 2 -j- 2 a . x. 

Multiplying both members of the equation by tri, the mass of 
the element, we have 

m r 2 = m r? -j- m a 2 -f- 2 a . ;;/ . x. ■ 

It is plain that for every element of the section A B a simi- 
lar equation could be obtained. Should the element ;// lie on 
the left of the plane of Y Z, the last term in the second member 
of the above equation would simply become negative. Con- 
ceiving now the whole mass to be cut by an indefinite num- 
ber of planes parallel to the section A B, and each element 
composing the sections thus formed to be connected by 
straight lines, with the points of intersection of the planes 
with the axes Yn and O E, we would obviously have for each 
-element of all the sections an equation similar to the one 



218 ELEMENTS OF MECHANICS. 

above. Summing up the equations thus derived, we have for 
the moment of inertia of the whole mass 

2 m r 2 = 2 m r* -\- a' 2 2 m -\- 2 a 2 m x ; 

or, denoting the whole mass by M — 2 m, and observing that 
the expression 2 m x, representing the algebraic sum of the 
moments of the elements of the mass with respect to the plane 
of Y Z, which passes through the centre of gravity, is equal to 
zero, section [83], this equation becomes 

2 m r 2 — 2 m r* + a 2 M, 

in which 2mr x 2 represents the moment of inertia of the mass, 
referred to the axis Yn through its centre of gravity, which 
by hypothesis is given, and 2 m r 2 its moment of inertia with 
respect to the parallel axis (?£,ata given distance a from the 
former. The second member of the equation being therefore 
known, the value of 2mr 2 is determined, and we conclude 
generally, that the moment of inertia of a mass with respect to 
any axis is equal to its moment of inertia with respect to a par- 
allel axis passing through its centre of gravity, plus the product of 
the mass by the square of the distance between the two axes. 

Corollary. — It follows hence that the moment of inertia, and 
hence the resistance to rotary motion, is the least possible, or a 
minimum, when the axis passes through the centre of gravity. 

The above equation may also be written under the form 

(2 m r 

or, making — j~- — k 2 , it may be written 

2mr' = M(k 2 + ay 

[190.] As we have already stated, the determination of 
the moment of inertia of bodies rotating upon an axis is gen- 
erally effected by the application of the integral calculus ; but 
many of the examples found below are original solutions by 



DYNAMICS. 219 

the simpler algebraic processes. The following will serve as 
illustrations. 

EXAMPLES. 

(a) Let it be required to find the moment of inertia of a mate- 
rial line of length I about an axis passing through one extremity 
(Fig. 103). 

Let m denote the area of a transverse section of the line, 
and suppose the length of an elementary portion of it to be 
denoted by h. Then mh will represent 
the mass of this element. If we further 
suppose the ce?itres of these elements to be 
at a distance from each other equal to h, 
and the first at a distance from the axis IG ' I03 ' 

equal to h, we shall have for the moment of inertia of the mass 
which the elements compose, when h is assumed indefinitely 
small, as also m, 

2 m r 2 = {m h) h* + (m h) (2 Jif + (m h) (3 lif + (in h) (4 lif + etc. 
= m h 3 (i -j- 4 + 9 + 16 -f- etc.). 

But the quantity contained within the parenthesis, summed 
up to n terms by the differential method of series, is equal to 

2 n 3 -f- 3 n 1 -\- n 



m h 3 
whence 2mr 2 = —p- (2 n* -J- 3 ;/ 2 + n). 

Since, also, the number of terms in the above series is 
equal to the number of elements, or equal to the whole length 
of the rod / divided by the length of an element h 7 we have 

/ 

n = h' 

Hence Smr > = — ( 2 -- 3 + ^ + _ 

or 2mr' = i m P + £ m P h + \ m lh\ 



220 ELEMENTS OF MECHANICS. 

Supposing h to be infinitely small, the sum of the ele- 
ments would constitute the whole line ; and since every por- 
tion of each element would be equally distant from the axis, 
we should have for the moment of inertia of the line 

2 m r~ == i m l\ 

or denoting the mass of the line by M, we have 

M = m /, 

and hence 2 m r~ — -J MP. 

If we recur to the equation 

'2 m r, 



2 m r 2 = M 



M 



'-+*)> 



and substitute for 2 m r* its value in the present example, 
\MT, and for a its value hi, we have 



12 mr' \ 

iMr = M[- w ±+ir), or iP 



i/ 2 = 



M ' 
and 2mr 1 i = ^ w Ml\ 

for the moment of inertia of the line about an axis passing 
through its centre of gravity or middle point. 

If the 2mr? had been first derived by the above method, 
the moment of inertia with respect to any parallel axis, 
whether passing through the extremity of the line or re- 
moved from it to a given distance, could likewise have been 
determined. 

(b) To determine the moment of inertia of a rectangular plate 
of insensible thickness, with respect to an axis lying in it and pass- 
ing through its centre of gravity. 

LetABDE (Fig. 104) be the plate, ' 

D 



and FG the axis passing through its F 

centre of gravity C parallel to A E. jB 

Conceiving the plate to be composed of Fig. 104. 

an indefinite number of physical lines parallel to A B, of 



DYNAMICS. 221 

equal length /, and whose masses are m, m v m v etc., we shall 
have for the moments of inertia of these lines, by the preced- 
ing example, 

p p p 

™—> Wi—, m *Tz' etc * 

But the moment of inertia of the whole mass is plainly equal to 
the sum of the moments of inertia of the elements m, m lt m 2 , etc. 
Hence, denoting the mass by M = m -\- m 1 -f- ;/z 2 -f- etc., we 
have 

/*,■-. . „/^ 

2 * 



2 m r* = — {in -j- m x -f- m 2 -f- etc.) = M 



If A E were assumed as the axis, then the moment of 
inertia could either be derived immediately, or by substitut- 
ing for 2 m r* its value above in the equation 



2 m r* = M 



\-M~+ a \ 



in which a would be equal to i I. We should thus obtain 

P 
3 

(c) When the axis is perpendicular to the plane of the rect- 
angle at C. 

Let mn be one of the elements of the plate, and C l its 
point of intersection with the axis F G. The moment of in- 
ertia of this element with respect to an axis passing through 
C x parallel to the axis drawn through C, perpendicular to the 
plane of the rectangle, will be (Ex. a) 

P 
m. — , 

12 

in which m denotes the mass and / the length of the element. 
Denoting the distance C C x by x, the moment of inertia of the 



222 



ELEMENTS OF MECHANICS. 



same element with respect to the axis passing- through C 
would be, section [189, Cor.], 



2 m ?"" = in - 



m r l 

m 



+ a' 



) 



m 

I T 1 



l * . > 

m — + m x . 

12 ' 



I 



J 



Similarly, if we take the nearest element to the point C, 
and denote the distance of its centre from C by h, we have 
for its moment of inertia 

m 1- m k* ; 



and for the next element, the distance of whose centre from 

C is 2 h, 

P 
m — +m(2ky; 

and for the third, * 

P 



etc. 



etc. 



But when h is very small, and equal the breadth of each ele- 
ment, the moment of inertia of one half the rectangle is equal 
to the sum of the moments of the elements. Hence 



r* — — (m + m -\- m + etc.) + m /i 2 \ 1 + 4 + 9 + etc - 



or 



2 m r 1 = 



M_ P_ 

2 12 



+m * l **+**+•], 



in which n denotes the number of elements composing one 

hall the rectangle, or — . 

The breadth of each element being equal to h, and the dis- 
tance of their centres apart being also equal to h } they would 



DYNAMICS. 223 

then be in contact and constitute the whole mass. In that 
case, however, we have 



m = Ih and n = 



2 



m Ik 

Substituting these values of m and n above, we have 

24 T 6 I r/1 3 ^ 6 pj? X ik )> 

24 ' 6 ( / 3 ' ^ / 2 T / j 

But when ^ becomes infinitely small, which is necessarily the 
case when the expression ;;/ — accurately represents the mo- 
ment of inertia of a single element, the above equation becomes 

24 ' 24 r 

Hence the moment of inertia of the whole rectangle will be 



12 



Smr,' = —3f\r.+ 1 r s , 



the axis passing through the centre of gravity. 

If the length of the rectangle be denoted by/, and its diago- 
nal by d, we have 

M=lJ>, or ~p-=f- 
Substituting this value, the above equation becomes 

2 m r ? = -iV M i r + f \ = tV Md\ 

It is easy to see that the moment of inertia of a rectanglar 
parallelopipedon, if its mass be denoted by M, and d be the 
diagonal of a face perpendicular to an axis passing through 
its centre of gravity, will be also 

2mr* = -^Md\ 



224 



ELEMENTS OE MECHANICS. 



id) The circumference .of a circle with a radius R, the axis 
being perpendicular to its plane, and passing through the centre. 

Assuming the unit mass ///, which is supposed by its mo- 
ment to give the unit of measure for the moment of inertia, 
to be equal to the radius of a circle, assumed as unity, and of 
the same physical character and magnitude as the given cir- 
cumference, we have for the whole mass M the expression 

M = 2nR\ 

and since every point in the circumference is equally distant 
from the axis of moments, we have 

2mr* = 2/uR 2 = R* 2 m = R 2 X M = R* X 2 n R = 2 ttR 3 . 

For the cylindric shell with the same radius, and altitude h y 
we should obviously have 

2mr* = R 2 + M= R 2 + 2 nRh = 2 ?t R 5 h. 



(e) A circular plate of insensible thickness with respect to an 
axis passing through its centre and perpendicular to its plane. 

Conceive the circle A B (Fig. 105) to be divided into a 
number of elementary rings m, m v m„ etc. 

Let the breadth of these rings, as well as the distance of 
their central circumferences apart, be 
denoted by r. The masses of these 
rings, when r is supposed to be infi- 
nitely small, will be equal to their cir- 
cumferences multiplied by r. That is, 
we shall have 

m — 2 n r o 7', 

m, = 2 n r, . r. 



i)i 



„ - 2 it r 2 . r, 
etc. etc., 



where r, r v r» etc., denote the radii 

of the several rings, having the common centre O. 




DYNAMICS. 225 

Hence the sum of the moments of inertia of the elements, 
equal to the moment of the whole mass, will be 

2 m r, 2 = 2 n r . r . i' 1 -\- 2 7r ?\ . r . r* -f~ 2 7tr 9 . r . r 2 2 -|~ etc. 
— 2 7t r (r 3 + ^i 3 + r* + etc.) 
= 2 7r r [r 3 + (2 r) 3 + (3 r) 3 + (4^.+ etc.] 
= 2 7T r 4 (1 + 8 + 27 + 64 + etc.). 

By the differential method of series, the quantity included 
in parenthesis summed up to n terms, is equal to 

i(* + 2 »■ + *■). 

Hence 2 m r? = VOL- 5 n * _j_ 2 ;/ 3 + *■ f , 

in which /z denotes the number of terms in the series, or the 
number of concentric elements composing the circle. 

Hence 
and 



But when r becomes infinitely small, which is necessarily the 
case when the expressions above for the values of the elements 
m, m v m„ etc., are accurately true, we have 

2.mr* = \n F? = iMR 2 

for the moment of inertia of the plate. 

If the axis pass through one extremity of the diameter of 
the circle, the moment of inertia of the plate with respect to 
this axis will be 

2 m r> = M j ^^- + R 1 1 = \ n R 4 + MR 1 







OB 

n = 

r 


R 






2 


ntr* 


_27tr i \R i 
4 \r* 


1 ^3 1 


R' 

r 1 


I 






= %ttR* + 


?rR 3 r + i 


7T 


RJr\ 



226 ELEMENTS OF MECHANICS. 

For the cylinder with R for the radius of its base, and of 
altitude h, we should plainly have 

2 m r* = i n R* . h = \ M R\ 

If the cylinder circumscribes a sphere, 

/i = 2R and 2mr?^nR\ 

(f) The moment of inertia of an annulus lying between the 
circumferences R and R t will obviously be equal to 

\n{R- r:\ 

(g) The circumference of a circle with respect to its diameter 
as an axis. 

Taking any small arc ns = m (Fig. 105); let r denote its 
distance from the axis A B, and VW— up = h equal its pro- 
jection upon the axis. When h becomes infinitely small, we 
shall have from the similar triangles m h and n sp the pro- 
portion 

Rh 
m : h : : ] m : mh : : R : r, or m = — , 

r 

and the moment of inertia of this element is equal to 

m r* = Rh r. 

If in like manner we denote by r, r„ r 2 , etc., the distances 
of the various elements from the axis A B, we have for the 
moment of inertia of the semi-circumference 

2 m r" = 2 R k r = R 2 h r = R \ hr + // , r x + h % r, + e t c . \ . 

But h being infinitely small, and the triangle nsp rectilinear, 
the quantity within the parenthesis is equal to the sum of all 
the trapezoids similar to vwsn\ that is, equal to the semi- 
circle A B R = i 7t R\ 

Hence 2 m r 2 = R .1 71 R 2 = A n R 3 



DYNAMICS. 227 

for the moment of inertia of the semi-circumference, and for 
the whole circumference 

2 m r; = it R\ 

(h) A circular plate of insensible thickness with respect to its 
diameter A B (Fig. 105) as an axis. 

The moment of the circumference of a circle, by the pre- 
ceding example, is 7tR 3 . If this circumference be supposed 
to have a breadth equal h, where h is indefinitely small, its 
moment becomes 

n R" h. 

But the whole circle may be supposed to be made up of an 
indefinite number of these concentric elements ; the radius of 
the first, nearest the centre of the circle, being h, of the sec- 
ond 2 h, of the third 3 h, etc. 

Hence the moment of inertia of the circle would be 

2-m-r* = 2 n R 3 h — n h 2 R 3 = it h\ h 3 + (2 h) 3 + (3 lif + etc. }, 
2mr? = 7t]?\\ -f-8 + 27 + etc.J = n h" \\{n* + 2 n 3 -f n 2 )}, 

in which n denotes the number of the elements. 
We have, therefore, 

R 
n = I> 

^ a nh K ( R 4 , R 3 R'\ ttR* , 7tR z k , n R 2 h % 
2mY? = — - \ -Ti + 2 -T3 + -rA = — - + — — - + 



4 ( h* ' h* ' #M 4 ' 2 ', 4 ' 

or, since ^ is infinitely small, 

2mr* = 1*1? = IM!?. 

The student can easily show that if the axis be taken tan- 
gent to the circle, its moment will be f n R\ 

(i) To find the moment of inertia of a spherical shell of insen- 
sible thickness with respect to an axis passing through the centre 
of the sphere. 



228 



ELEMENTS OE MECHANICS. 



Let A B (Fig. 106) be a section of the sphere, and suppose 
the spherical shell to be cut by a system 
of parallel planes perpendicular to the 
axis CD, and at equal central distances 
apart, which distance denote by d. The 
surface will then be cut into a number 
of equal zones, as shown in the geometry ; 
and if d becomes very small, and we de- 
note by R, R lf R„ etc., the radii of the 
successive zones, commencing at the 
equatorial zone, the moment of inertia of 
each, since at the limit d becomes the breadth of the zone, 
will be (Ex. d) 



4^~ 







\ 




ir \ 


1 d 


*a \ 




s i \ 




f Jb 



Fig. io6. 



For the first, 
For the second, 
For the third, 
etc. 



2 7tRdX R*\ 

2 7t Rdx R*; 

2tc Rdx R?\ 

etc. 



But R 2 = (R + d)(R-d) = R 1 - d 2 X i, 

R 2 = (R + 2 d) (R - 2 d) = R 2 — d 2 X 4, etc., to n terms. 

Hence 2 m r 2 = 2 n R d \ R 2 -\- R t 2 -j- R 2 -\- etc. to n terms \ 

= 2 7tRd\nR 2 — d 2 (i -\-4-\-g-\- etc. to n terms) \ , 



or 



2mr 2 = 27t Rd\ nR 2 



d'~ 



2 n 



= 2 7tR^dnR 2 - d 3 i^~^ 

But n d = R, 

R z 
whence, also, d % — — 

n* 

By substituting these values, and making n — 00 



n 2 + n \ \ 



m r 



27T R\\R'\ =4 t 7tR 4 ; 



DYNAMICS. 229 

and for the whole spherical surface, 

2mr* = i7rR i = f M R\ 

(J) To find the moment of inertia of the solid sphere about its 
diameter. 

By the preceding example we have, for the moment of 
inertia of the spherical shell of radius R, 

2mr* = %7t R\ 

But the solid sphere may be conceived to be composed of 
concentric shells in contact. Let R iy supposed to be very 
small, denote the radius of the first spheric shell, commencing" 
at the centre of the sphere ; R 2 = 2 R x that of the second ; 
^3=3^, the radius of the third ; and so on to R, the radius 
of the sphere. Let R t also equal the thickness of each shell. 
They will then lie in contact, and constitute the sphere. 
The moment of inertia of the first will be 

of the second, f n R* x R x ; 

etc. 
Hence the moment of the whole sphere will be 

2 m r? = I n R x {R* -f R^ -f R^ -f etc. to n terms say}, 
orSwr^f 7tR 1 {R l 4 + R l \2 i + R 1 i .3 i +R 1 \4 i -\- eta n terms}, 

2mr? — \7tR*\\ + 2 4 + 3 4 + etc. n terms} 
n . ( n* , n* . n z n ) 

3 1 5 2 l 3 30 ) 

But nR, = R, or R* = — ; 

and by substitution, and supposing n to become infinite, 

2 m r* = f n R 6 X £ = A * R * = t ^^- 




23O ELEMENTS OF MECHANICS. 

{k) Let the student find the moment of in- 
ertia of the line A C (Fig. 107) revolving about A 
A B as an axis. 

Suggestion. — Let A C be supposed to be uni- 
formly projected upon B C, in which case the 
moment of each of its elements will be the same g 
as before. Fig. 107. 

Centre of Gyration. 

[191,] The centre of gyration of a body is that point at 
which, if the whole mass be supposed to be collected, its 
moment of inertia with respect to a given axis will remain un- 
changed. The distance of this point from the given axis is 
called the radius of gyration. 

Denoting the mass by M, and the radius of gyration by 
K v we shall have by the definition 

that is to say, the radius of gyration is equal to the square root 
of the quotient obtained by dividing the moment of inertia of the 
body, with respect to the given axis, by its mass. 

The mass being supposed to remain constant, the radius of 
gyration will be a minimum when the moment of inertia is a 
minimum; that is, section [189, Cor.], when the axis passes 
through the centre of gravity of the body. It is then called 
the principal radius of gyration, which will be denoted by K. 



Principle of D'Alembert. 



[192.] In considering the effects of forces acting upon a 
system of bodies, or parts of the same body, when they are 
connected amongst themselves by some law of mutual action, 
it becomes frequently very advantageous to refer to a prin- 
ciple introduced into mechanics by D'Alembert, which may 
be regarded almost as a dynamical axiom. 



DYNAMICS. 231 

The following is in substance its general statement : 
If there be a system of bodies A, B, C, etc., which, in virtue of 
the impressed forces ff,f 2 , etc., would, if perfectly free to move, 
receive, in the first instant of time, respectively the velocities a, b, c, 
etc., but which, on account of their mutual connection, receive in- 
stead the velocities v, v v v„ etc., then will the forces which are due 
to these latter velocities, when reversed in direction and taken in 
connection with the impressed forces, produce equilibrium in the 
system. 

For the forces due to the velocities v, v v v. 2 , etc., or which 
give rise to these velocities, being the resultant forces which 
arise from compounding the impressed forces /, /„ f, etc., 
with all the internal forces of the system, may be considered 
as the only effective forces which act upon it, and therefore 
must, if reversed in direction and taken in connection with 
the impressed forces, and the internal forces of the system 
which produced them, necessarily restore the equilibrium 
previously existing : static equilibrium if the body or system 
were at rest before the application of the impressed forces, 
and dynamic equilibrium, or uniform motion, if the system 
were already in motion. 

As an illustration of the principle just enunciated let us 
suppose A, B, and C (Fig. 108) to be three bodies, and let A m 
and A n represent the forces acting upon % g # 

A through the influence of B and C in 
any manner. These two influences com- 
pounded give a resultant A R — R. If 
we now suppose the force Af to be im- 
pressed upon A simultaneously with the 
combined action of B and C, equivalent 
to A R, we have a resultant A R. as the 

re r , • • FlG " Io8 ' 

only effective force of the system, giving 

rise to a velocity v, say, for A. But it is plain that if A R v the 
force due to this velocity, be reversed in direction it will pro- 
duce equilibrium. 

Or again, if we suppose the velocity a which the im- 




232 ELEMENTS OF MECHANICS. 

pressed force / is assumed to impart to A when perfectly 
free to move, to be resolved into two component velocities, 
one of which is the velocity v actually received, and the other 
the velocity b in the direction A F, then, since A F is parallel 
and equal to R 1 f t which is parallel and equal to A R, it must 
also be equal to A R, and the force due to the velocity b or 
A F will hold in equilibrium A R, representing the effect of 
the internal forces of the system ; that is to say, that com- 
ponent of the impressed velocity which is not the velocity 
actually received, is held in equilibrium by the internal forces 
of the system. Hence, as before, if the force due to the 
actual velocity be reversed in direction, there will be equi- 
librium. 

[193.] We shall now give several examples of the applica- 
tion of D'Alembert's principle that will further elucidate the 
subject. 

Example i. — Let it be required to determine the acceleration 
of two bodies W and W x (Fig. 109) placed upon opposite inclined 
planes, and connected with each other by means of a flexible cord 
passing over the pulley C. 

Denoting the lengths of the planes by / and l v and the 
common altitude by h, the impressed 
force on W to produce motion [21] 

will be represented by — j-, and that 

on JV t , since its motion will be in an op- 

posite or negative direction, by — — -, — . Fig. 109. 

If we now denote the effective forces on Wand W l — that is 
to say, the forces imparting to them their actual velocities — 
by P and P r respectively, then, by D'Alembert's principle, 
these forces reversed will, in connection with the impressed 
forces and the mutual action of W and W x upon each other, 
produce equilibrium. Hence we shall have 

^_P_^_P i = 0) or P+Pi = E±_E^. 




DYNAMICS. 233 

But representing by / and /, the accelerations P and P x 
are capable of imparting to W and VV X respectively in one 
second, we have, since two constant forces are to each other 
as the accelerations they are capable of producing in the same 
mass, in the same time, 

P:W::f.g and P, : W r .: f, : g- 
Hence P=^; 

cr 

and since the accelerations of the two connected weights 
must be equal, 

W f W f 
/ = /, and /> = ^IiZl = J2jZ. 

g g 

Substituting these values of P and P, above, we have 

Wf WJ _ Wh W\h 
g + g\~ I ~~k~' 

f wh wj r 

1 ~ /, " , ^ 

w+w x r 

J 



whence 



ior the common acceleration of the two bodies. 

Case 1. — If both planes be supposed to have the same in- 
clination, then / = /, ; and if we further suppose both planes to 
revolve upon the point C until they coincide with h, then 
J = L — h, and the formula last obtained becomes 



/ 






for the acceleration of two weights suspended over a pulley. 

The value of / being constant, the acceleration will be 
uniform ; and since the weights may be made to differ from 
<each other by a very small amount, the acceleration / may 



234 ELEMENTS OF MECHANICS. 

become as small as we choose, and the velocity of the weights 
so small that the value of / may be experimentally deter- 
mined. 

This is effected by means of Atwood's machine, to which 
the above formula applies, and for an explanation of which, 
and the method of conducting the necessary experiments, the 
student is referred to some elementary work on physics. 

Having thus determined the value of /, we have 



\ W- w. 



:!/■ 



and the value of g, the acceleration produced by gravity in 
one second, is also determined independently of the methods 
for determining the same quantity in section [144]. 

Case 2. — If one of the planes only should become vertical 
as /, we should have / = //, and the general formula for the 
value of /becomes 



1 w+w- 



\g' 



Case 3. — If one of the planes, as /, becomes vertical, and 
the other, /„ horizontal, then I — h and /, = 00 , and we shall 
have 

W 
f 



If now W = W„ we have 

as should obviously be the case, since the gravity of a single 
weight would be moving the two weights combined, over- 
coming simply the inertia of both. 

If, in the cases considered above, we wish to determine 
either the velocity acquired in a given time / or the space 
passed over in the same time, it will be necessary simply to 



DYNAMICS. 



235 



substitute the value of f determined in each case in the for- 
mulas for the space and velocity where the acceleration for 
one second is denoted by/. 
These formulas are 



S = i/f 



and 



V=ft. 



EXAMPLE 2. — A given weight ^(Fig. no) Jianging vertically 
is made to draw another given weight W x up an inclined plane l x 
of a given altitude h ; it is required to find the length of the plane 
l x when the time of ascent is the least possible, or a minimum. 

In the formula 5 = iff 2 , when 
S = ^ we have 

h - kft\ 

in which f is the acceleration for 

one second. But the present ex- Fig. ho. 

ample corresponds with Case 2 of the preceding-, in which 

w- w/y 




f 



WAr W x 



Hence, by substitution, 

2l l (W+ W,) 



'-?■- 



2 (JV+JV l ) 



g [w-w x j 



- X 



W- w x — 



2 (iy+W 1 ) 



X 



/, 



IV I. - W,h 



Since the factor ! is a constant quantity, f and 



therefore t will be the least possible when -y^j — L ~fj7 _ T is the 

least possible, or a minimum. 

The question is therefore resolved into the following ; viz.,. 

r 



What value of /, will render 



Wl x - W, h 



a minimum r 



236 ELEMENTS OF MECHANICS. 

This value of l x may be determined by the following alge- 
braic process, without having recourse to the differential cal- 
culus, usually employed in questions of maxima and minima. 
Thus, let W x h — q, and put 



Wl x -q 
a minimum, and we have 

/;_ Wul x + uq = 0. 

Wu 
Let /, = x -j- , and by substitution in the preceding 

equation we have 

W 2 u 2 W 2 u 2 

x 2 + Wu x 4- — Wu x f- u q = o, 

4 2 

2 W 2 u 2 
or x — -4- uq = o, 

4 ^ * 

which solved with respect to u gives 



«-££ + /if- + ±£ 

from which it is evident that // will be a minimum when 
jr = o. But if x = o, then 

4? 

* ~ W 2 ' or M ~ °' 

the latter value, since it would render l x — o, being inadmis- 
sible. Hence 

4q Wu W 4q q 

U = W and K = * + ~T = ° + T ' W = 2 W 

or, replacing q by its value, 

_2W\k 

1 ~ W' 1 
the required length of the plane. 



DYNAMICS. 



237 



Note. — The value of l x determined above is the same as 
that derived by the method of the calculus, thus : 



K 



du 
dL 



- wi x -W x h 
(Wl x - W 1 h)2l 1 



a minimum ; 
L x W 



Hence 



( wi x - w x hy 
2 wi? - 2 w x i x h- wi; 

Wl* -2W x l x h = o, 

Wl x =2W t h, 

2 W x h 



o. 



o, 



/, = 



W 



-, as before. 



Example 3. — As a third illustration of UAlemberfs prin- 
ciple, let it be required to determine the motions of two weights 
W and W x (Fig. in) attached to a wheel and axle, as shozvn in 
the diagram. 

The impressed forces are plainly the weights IV and — W x , 
the weight W being supposed to pre- 
ponderate. Denoting the effective forces 
by P and P x respectively, the forces pro- 
ducing equilibrium in the system will be 
(W-P)zmd(- W X -P X ) = -{W X + P X ). 
But these forces act with the levers R 
and r respectively. Hence, on the prin- 
ciples of the lever, 



fW- P).R-(W 1 + P 1 )r = o. 




Fig. hi. 



Representing by /and /, respectively 
the accelerations of J^and W x produced by P and P v we have 



P: W::f:g 



:.P 



and P x : W x \\ f x \ g 

Wf 



238 ELEMENTS OE MECHANICS. 

and since the accelerations of the weights are as the lever- 
arms R and r, 

/:/,:: #:r, and A=^' 

Hence /> = -f = _/, 

and, by substitution above, 

or j? fT- ^-^ == r ^ + ^~; 

^ £^ , 

( WR* —W,Rr\ 
Whence / = l- ^'+^r' -l^ 

a formula for the acceleration of the descending weight. 

r 
If this value of / be substituted in the equation /j = -77/, 

we have 

WR r-W.r * 

WR' + W x r 



f* — ( U/ I?* I U/ J* X 8 



for the ascending weight. 

The denominators of these values for / and f x being the 
same, we have 



/ 


WR'- W x Rr 


R(WR- 


W x r) R 


/r 


WR r-Wy ~ 


- r {WR- 


W x r)~ r 



as should be the case, since the velocities of the two weights 
would obviously be as the radii of the wheel and axle. 

To find the velocity acquired in a given time /, or the 
space accomplished during the same interval, by the weight 
W, the value of/ found above may be substituted in the gen- 
eral equations 

V = ft and S = %ft\ 



DYNAMICS. 



'59 



[194.] It will be seen from the examples we have already 
given of the application of the general principle of D'Alem- 
bert, that it depends entirely upon our ability to find expres- 
sions for the impressed and the effective forces of the system. 

Some further applications of the principle will be found in 
portions of the text farther on. 



The Compound Pendulum. 

[195.] In discussing the laws of the vibration of the simple 
pendulum, and the uses to which it could be applied, it was 
conceived to have only an ideal existence, and its whole mass 
was supposed to be concentrated in a material point, vibrat- 
ing about a fixed centre, with which it was connected by 
means of a mathematical line, inflexible and inextensible. 

The object of the present article is to determine some of 
the questions involved in supposing the pendulous mass to 
have a physical character, as must necessarily be the case in 
conducting all the experiments in which its use is of import- 
ance. 

Let A B (Fig. 112) represent the section of the pendulum 
made by a plane coinciding with the plane of the paper ; P, 
the point at which an axis of suspension perpendicular to the 
plane of the paper pierces the section A B ; and w, a material 
particle at a distance r from the point of suspension P. If we 
now take a particle ;;/, at the 
unit's distance from P, and de- 
note by go its angular velocity 
at the moment the pendulum 
occupies the position A B in its 
descent in the direction of the 
curve CD, we shall have for the 
velocity of the particle m, at the 
same instant of time, r 00. 

If we further suppose the in- 
crement of the velocity of m. for an instant of time /. to be de- 




FlG. 



240 ELEMENTS OE MECHANICS. 

noted by co v its velocity would then be (go -f- a?,), and the velo- 
city of the particle «, would be r go -\- r go v the quantity r ^ 
representing the increment in the velocity of m during the 
same instant of time. 

This value r co^ it should be observed, is the actual incre- 
ment of velocity of the particle m in its connection with the 
whole system of particles composing the pendulum. If, how- 
ever, the points m and m x were supposed to have no physical 
connection with each other or with the other particles of the 
mass, and to be free to vibrate as simple pendulums, it is evi- 
dent that the angular velocity of m x , section [138], would be 
the greater. 

On the other hand, a particle taken at a greater distance 
from P than m would have a less angular velocity than m. 
Hence in the pendulum all those particles on the same side of 
P with m, and lying between P and m, would tend to ac- 
celerate the motion of the latter, and all those at a greater 
distance from P than m to retard its motion. There must 
therefore be some particle in the mass with reference to 
which the forces of acceleration and retardation due to the 
remaining particles, exactly counterbalance each other, which 
particle would therefore vibrate in the same time as if it con- 
stituted a simple pendulum suspended at P. 

Representing the weight of the particle m by the line 
mg = g, and resolving this weight into its components mp 
and mf, the former will be destroyed by the reaction of the 
point of suspension P, and the latter, mf wili represent the 
accelerating force acting upon the particle. 

Denoting the angle Pm n by S, we have 

Pm n — fmg = S and mf — g cos S. 

In this equation g denotes the weight of the particle or force 
of gravity, and g cos S the force causing the vibration ; hence, 
since the accelerations due to any two forces in the same time 
are as the forces [4], if we suppose g to denote the accelera- 
tion of gravity for the unit of time, 1 second, ^-cos 3 will be 
the acceleration due to the tangential force for the same time, 



DYNAMICS. 241 

and gcosd .t x the acceleration for an indefinitely short time 
t 1 [4, Cor. 3]. Multiplying by m the mass of the given particle, 
we have 

mg cos 8 t x 

for the quantity of motion that would be imparted to a free 
particle m by the accelerative force g<zos 8 in time t\. 

In like manner, r go x being the increment of velocity actually 
imparted to m on account of its connection with the system, 
we have for the real or effective quantity of motion of the par- 
ticle, mr oo j. 

Since every other particle of the mass would give rise to 
expressions similar to the above, we have 

2 mg cos 6 t x 

for the quantity of motion which is the measure of the im- 
pressed forces, and 

2 m r oo x 

for the quantity of motion which really obtains as the effect 
of the resultant or effective forces of the system. 
But on the principle of moments we have 

r . mg cos S t x 

for the moment of m due to the impressed force, and 

r .mr oo 1 = m r 2 go 1 

lor the moment due to the effective force upon the same par- 
ticle. 

Hence by D'Alembert's principle, reversing the effective 
forces of the system, there results 

2 mgr cos d t x — 2 m r 2 go, = o, 

and 2 mgr cos §t x = ^mr 2 go x ; 

or, since g and t x are constant factors common to all the terms 
composing the first member of the equation, and go, a factor 



242 ELEMENTS OF MECHANICS. 

common to those of the second member, this equation be- 
comes 

gt x 2 mr cos d = gd x 2 m r 2 , 

. gj, 2 m r cos d 

and — - = ^ — • g. 

t x 2, mr d 

Conceiving now a vertical plane to be passed through the 
axis of suspension perpendicular to the plane of the section 
A Bj and denoting the distance of the particle m from this 
plane, equal to m n, by y, we shall have 

mn = y = r cos <?, 

and the equation above becomes 

oo x 2 m y 

l x ~ =g '^n~?' 

in which the expression 2 my denotes the algebraic sum of 
the moments of the particles of the mass with respect to the 
plane passing vertically through the axis. 

Hence if we denote by j\ the distance of the centre of 
gravity of the pendulum from the same plane, and by M its 
mass, we shall have, section [8o], 



Hence 



If now a plane be passed through the centre of gravity of 
the pendulum parallel to the section A B } and we denote by 
P x the point at which the axis of suspension pierces this plane, 
by M 1 the position of the centre of gravity, and by n x the 
point at which the perpendicular from the centre of gravity 
pierces the first plane, there will be formed an angle P 1 M 1 n v 

Denoting this angle by d„ and the distance from the axis 
of suspension to the centre of gravity, equal to P x M v by a, we 
shall in like manner as before, in finding the value of y, have 

y x — M x n x — P x M x cos 6 X — a cos d x . 





2 my = M y x . 


G0 X 


2 m y M ' y x 
~~ g 2 mr* ~~ * 2 m r 2 



D YNAMICS 243 

The last equation therefore becomes 

gd 1 Ma cos d, 
!~ =g ~ 2 m? 

But we have also, section [189, Cor.], 

2tnr* = M \k 2 + a 3 \. 



Hence 



oo l MacosS^ gacosd 1 

77 = g W(¥'+~a 1 ) = k 2 + a 2 ' 



, ga cos d x 

and ">>--¥+!?'» 

in which ao l denotes the increment of angular velocity for the 
time t x indefinitely small. 

If now we have a simple pendulum, of length /, mak- 
ing with the horizon the angle tf, = P, m x n v the equation 

— - = g - becomes for a single material particle, denoting 

the increment of angular velocity in the time t x by go„ 







*>* my gy gy 

t, ~ g mr 2 ~ r 2 ~ P 


But 




y = /COS #„ 


and hence 


^1 


gcosd, g 


— 1 , &? 2 — ^ CO! 



in which, as already supposed, g? 2 is the increment of angular 
velocity for the simple pendulum, having the same angle of 
inclination ^ with the horizon, as the compound pendulum. 

If we make the increments of angular velocity equal to 
each other, and suppose the two pendulums to start from rest 
at the same elevation, these increments for every instant of 
time t x will be equal, and hence the two pendulums would 
vibrate in the same time. 



244 ELEMENTS CF MECHANICS. 

To determine, therefore, the length / of the simple pen- 
dulum that will perform its vibrations in the same time as the 
compound pendulum, we have only to make 



gd 1 = g? 2 , or 


gacosd^ £-cos 


*. 


•A- 


je + ^ '*>- I 






, k * + a * 

and / = ! 

a 


= 




k -f- a I 



the required length. 

2 m r 2 
Since k? = — ^r- [189, Cor.], it becomes necessary, in or- 
der to apply the above formula, to know the mass of the pen- 
dulum, its moment of inertia with respect to an axis through 
its centre of gravity parallel to the axis of suspension, and the 
distance of these two axes apart, denoted by a. 

If we replace ~2mr? by its value 2mr 2 — a 2 M, we shall 
have 

2 m r 2 2 m r* - a" M 



1 = -luT- +a = JJ- + a = 



Ma ' . Ma ' Ma ' 

in which 2 m r 2 is the moment of inertia of the pendulous 
mass with respect to the axis of suspension. 

[196.] Now let CD (Fig. 113) represent a section of the 
compound pendulum by a vertical plane passing 
through its centre of gravity M 1 and perpendicu- 
lar to the axis of suspension passing through P v 
Draw P 1 M 1J and produce it to 0, taking off P 1 o, 
equal to the length of the simple pendulum which 
will vibrate in the same time as the compound 
pendulum. 

Through draw a line parallel to the axis of 
suspension. This line is called the axis of oscilla- Fig. 113. 
tion, and the points P, and the centres of suspension and oscilla- 
tion respectively. 




DYNAMICS. 245 

Problem i. — To find the distance of the centre of oscillation 
of a semicircular plate of insensible thickness from the centre of 
the circle when the diameter becomes the axis of suspension. 

We have by the preceding section for the length of the 
simple pendulum which will vibrate in the same time as the 
semicircle, considered as a pendulous mass, the formula 

2 m r 2 
~Wa~~' 

But in this case we have, section [190, k~], 

2mr 2 = i7r R\ 

4R 

and, section [80, c], a = 

Hence, since the area representing the mass of the semicircle 
is equal to \n R* = M, we have, by substituting the above 
values, 

2mr i^R* 3 



/ = 



Ma 1 2 4., 
3* 



X=^7tR = .& 9 R 9 



the required distance or length of the equivalent simple pen- 
dulum. . 

Problem 2. — To find the distance between the axis of sus- 
pension and the centre of oscillation of a slender rod or material 
line suspended at its extremity. 

Denoting the length of the line by l v we have in this case 
[190, a] 

2m't* = iMI* and *=JrA- 

Whence /^^-i^U*/ 

l ~ Ma ~ M.\l~ 3 " 

the distance required, or the length of the equivalent sim- 
ple pendulum. 



246 ELEMENTS OF MECHANICS. 

PROBLEM 3. — To find the value of I in the case of the sphere 
suspended by, and vibrating about, a tangent line. 

The moment of inertia of the sphere about the tangent is, 
section [189, Cor.], 

{2 m r a 
—if ■-+«' 

But the value of 2mr?, the moment of inertia of the 
sphere about its diameter, as heretofore found [190,/], is 







2 


mr? = %MR\ 




and the distance 




a = R. 






.-. 2 


mr* = 


%MR 2 + MR 2 -- 


-\MK 


Hence 


1 = 


2 mr 
Ma 


= \ =1* 


10 ' 



in which D is the diameter of the sphere. 

The student can in like manner find the length of the 
equivalent simple pendulum in any case where the quantities 
2mr* a and M can be determined, or have known values. 

PROBLEM 4. — A circular plate of thickness h (Fig. 114) and 
radius R vibrates about the point P at a distance 
Po = a. It is required to find the length of the 
equivalent simple pendulum, the plate being supposed 
to vibrate edgewise. 

We leave the solution of this example to the 
student. It may be observed, however, that / will 
be greater than Po or a ; for if the arc m n be de- 
scribed upon the plate from Pas a centre, it is ob- 7n &W) n 
vious that the angular velocity of will be retarded _ 
by all the particles of the mass below the arc ;;/ n, 
and accelerated by all those above it. Hence the retardation 
due to a greater quantity of matter ; also, at a greater dis- 
tance from the centre of moments P, will exceed the accelera- 



D YNAMICS. 247 

tion, and the vibration will require a longer time than a 
simple pendulum of length a. We conclude, therefore, that 
/ > a. The true result will be 

1 2 a 
If R = 2 inches and a = 10 inches, 

/ '== 10 + \ — io£ inches, 
equal -§- inch in excess of a. 

[197.] The axes of oscillation and suspension are reciprocal; 
in other words, if the axis of oscillation becomes the axis of sus- 
pension, the latter will become the former. 

For let / be the length of the simple pendulum which will 
vibrate in the same time as the compound pendulum whose 
section (Fig. 113) is represented by CD. We shall then have, 
from what precedes [195], 

l = T + ' = & + **■> 

or, taking off M x equal to p M , we shall have 

l=P } o. 

If we now make the centre of suspension, and the axis 
of oscillation the axis of suspension, the length of the simple 
pendulum which will vibrate in the same time as the com- 
pound pendulum will be 

But since a x = M, — p M , we have 



/, = P x M l + -p-jyr, or /, =: /. 



248 ELEMENTS OF MECHANICS. 

The lengtJis of the corresponding simple pendulums being there- 
fore the same, the times of vibration of the compound pendulum 
about the axes of oscillation and suspension will be equal. 

[198.] Conversely, if the times of vibration of a compound 
pendulum are the same about tivo axes unequally distant from the 
centre of gravity of the pendulum, which point is included in the 
plane of the axes, then will the distance between the axes be equal 
to the length of a simple pendulum which will vibrate in the same 
time. 

Referring again to Fig. 113, let P, and become alter- 
nately the centres of suspension, and denote by / and l x the 
values of the corresponding simple pendulums. We shall then 
have 

ff l? 

I = \- a and /, = — 4- a.. 

a ' a 1 ' 

But since by supposition the times of vibration of the com- 
pound pendulums about the two axes are equal, the times of 
vibrations of the equivalent simple pendulums corresponding 
to them must be equal, and hence their lengths must also be 
equal ; or 

'= A, 

£* k" 

and hence — + a = — -\~ a,. 

a ' a x ' 

Solving this equation with regard to a„ we find 

a. = a, or a, = 

a 

But the distance between the axis is equal to 

P x — a + a x ; 

or, replacing a x by its second value, since the value a t = a is 
excluded by hypothesis, the mass not being symmetrical in 
form, we have 

P t o = a + — =1, 

1 a 

as enunciated. 



DYNAMICS. 249 

For certain symmetrical bodies we might also have 

a — a„ 
as will be easily seen. 

[199.] When discussing the simple pendulum, section [143], 
a method was given for determining the length of the seconds 
pendulum for any latitude. The foregoing section presents 
also a second experimental method of solving the same prob- 
lem. A straight bar of iron (Fig. 115) is made to 
vibrate with- the least possible friction upon two knife- rV 
edge axes, a and fr, resting upon horizontal plates of 
polished agate. 

The bar is so constructed that the times on either 
axis will be nearly equal. The time of each vibration 
is ascertained in the usual manner by dividing the 
whole time of any number of vibrations through a 
very small arc by the counted number of vibrations 
in the given time. Map 

One end of the bar is then filed away until the ** 
single vibrations upon the two axes become accu- 1G ' II5 * 
rately equal in time, and this time is noted. Calling the time 
/,, the distance a b between the axes /„ we have l x accurately 
measured by using a scale of equal parts for the length of the 
simple pendulum whose time of vibration is t v Now let 
/ denote the length of the required seconds pendulum. 
Then, since the times of vibration of simple pendulums in 
the same latitude are as the square roots of their lengths, we 
have 

1 : /, :: VT\ VT V 

VT l 

Hence Vl— — -, l=-h', 

which, since /, and t x have been found, makes known the value 
of /, the length of the seconds pendulum. 

The time of the vibration of a simple pendulum through a 

very small arc being t = n y — , if we make / equal one sec- 



250 



ELEMENTS OF MECHANICS. 



ond, / becomes the length of the seconds pendulum, found as 
above. Hence 

for the acceleration due to gravity in one second. 



Centre of Percussion. 

[200.] If an obstacle be struck by a rod of some weight, 
the hand with which it is held will, in general, receive a 
shock. If the point of the rod which comes in contact 
with the resisting object is near its farther extremity, the 
shock received by the hand will be in a downward direction ; 
but if near the hand, the shock will be upward. It follows 
that there is a point between the two which, meeting the ob- 
stacle, will impart no shock to the hand. This point, at 
which the full effect of a blow may be imparted to the resist- 
ing obstacle without any portion of it being lost upon the 
hand, or an axis of motion in its stead, is called the centre of 
percussion. 

The object of this article is to determine its distance, in the 
case of any rotating body, from the axis of motion. 

It is obvious that the whole force of the rotating body is 
received, when the obstacle arrests at once the moving forces 
of all the elementary masses of the body ; that is, when there 
is no resultant tendency to motion arrested by the resistance 
of the axis. The sum of the moving forces, or momenta of 
the elementary masses, will therefore be the measure of the 
whole moving force of the entire mass acting at its centre of 
percussion. 

Let m (Fig. 1 16) be one of 
the elementary masses of the 
body lying in a section A B, 
made by a plane perpendicu- 
lar to the axis Z o at o, and 
passing through the centre 
of gravity of the body at G. 




Fig. 116. 



DYNAMICS. 251 

Suppose the axis of X to pass through and G, and the 
elements of the body to be referred to the two co-ordinate 
planes X Z and YZ at right angles to each other, and passing 
through the axis oZ. 

Let F be the moving force of the element m acting per- 
pendicularly to m, the line joining m with the axis; and 
resolve the force F into the two components f and f v re- 
spectively parallel to the planes X Z and YZ. 

Denoting the angles Fmf=mo Y and Fmf x = moX 
respectively by y and /?, the distance m by r, and the angu- 
lar velocity of the body by a, there will result, in conformity 
with the notation adopted, the following : 

Velocity of m equal r a ; 

Moving force of m equal mra. 
Likewise /= Fcos y = mra cos y; 

f x == Fcos /3 = mra cos /?. 
But denoting the co-ordinates of m by x and y y 

y 1 x 

cos y = — and cos p = — . 
r ' r 

Whence/ = m ay, component of F parallel to the plane of XZ; 

f^ — max, " " " " YZ. 

If now we conceive the body to be cut by a series of planes 
parallel to the plane of the section A B, and indefinite in num- 
ber, it is obvious that like expressions could be found for all 
the elements of the body which will lie in these planes, and 
there would result 

2 m ay = a 2 my 

for the resultant of all the components parallel to XZ, 

and 2 max = a 2 mx 

for the resultant of all the components parallel to YZ. 



252 ELEMENTS OE MECHANICS. 

But if X } and Y x denote the co-ordinates of the centre of 
gravity of the body, and M the whole mass, we have 

2 m x = MX X and 2 my = M Y v 

Hence a 2 m x = a M X v 

But the centre of gravity lying in the plane of X Z, we have 

y, = o, 

and hence 2 my = o and a2my = o. 

The total resultant rotary force of the system of particles 
is therefore represented by 

a 2 m x = a MX i 

acting parallel to the plane of YZ, and hence perpendicularly 
to the plane of X Z at some point, d say, which will obviously 
be the centre of percussion of the body, since, if the resultant 
force a M X x were reversed in direction at this point, it would 
check the motion of the whole system of particles and pro- 
duce rest. If d denote the distance of the conceived point 
d from the axis, the moment of the resultant force aMX l 
would be 

aMX t d; 

and since this moment must be equal to the sum of the mo- 
ments of the moving elements with the same angular velocity 
a, equal to a2mr'\ section [187], we have 

aMX x d— a2mr\ 



° r d = ~WX~ = ~WaT = 7 ' section [ x 95]. 

Hence we conclude that the centre of percussion is at the same 
distance from the axis as the centre of oscillation. 

If, therefore, as supposed, the impact be in a direction per- 
pendicular to the plane passing through the axis and centre of 



DYNAMICS. 253 

gravity, the centres of oscillation and percussion will both lie in 
the same line parallel to the axis of suspension. 

In order, however, that the axis may not suffer a twist, 
even when not receiving a direct shock, the plane passing- 
through the centre of gravity and perpendicular to the axis 
must divide the body symmetrically ; in which case, also, the 
centres of oscillation and percussion will coincide. 

[201.] If now we suppose the mass itself not to move, 
but instead to receive an impact at the centre of percussion, it 
is obvious that the axis which before received no shock from 
the collision with a resisting object will in this case likewise 
be exempt from any shock ; and hence, not having imparted 
to it for the instant any tendency to motion, will remain for 
that instant stationary ; so that the whole mass in its initial 
motion will tend to turn spontaneously about an imaginary 
axis, which may be supposed to take the place of the real axis. 

This is called the axis of spontaneous rotation, of which we 
shall treat in some of the following sections. 



On the Motion of the Centre of Gravity, and the Position 
of the Centre of Spontaneous Rotation. 

[202.] In considering the motion of the centre of gravity 
of a body or system, we shall first take a system of material 
particles m, in iy m v etc., supposed to be entirely free to move, 
and unconnected by any law of reciprocal action of whatever 
kind. Let these particles be supposed to move in parallel 
directions with the uniform velocities v, v v v v etc., due to any 
given impressed forces. Conceive a plane to be passed through 
the centre of gravity of the system parallel to the common 
direction of the motion of the particles. 

It is obvious, section [80], that at the beginning of motion 
the algebraic sum of the moments of the elements m, m v m„ 
etc., with respect to this plane will be equal to zero ; and since 
the motion of each element is parallel to the plane, preserving 



254 ELEMENTS OF MECHANICS. 

a uniform distance from it, this moment will be equal to zero 
at every instant during the motion. 

It follows, therefore, that the centre of gravity will always 
lie in this plane, and hence must move in it. In like manner 
it must move in any other plane passing through it, parallel 
to the given direction of the motion, and hence in the line of 
intersection of the two planes. In other words, the centre of 
gravity of the system will move in a straight line parallel to the 
given directions of the elements m, m x , m„ etc. 

If we now conceive a plane to be passed perpendicular to 
the direction of the motion, and denote by d, d x , d„ etc., the 
distances of m, m x , m„ etc., from this plane at the commence- 
ment of motion, their distances at the expiration of the time t 
will be, since the velocity of each is supposed to be uniform, 
d -f- v t, d x -f- v x t y d^-{- v 7 t, etc. ; and if a and x be the distances 
of the centre of gravity of the system from the plane, at the 
beginning and expiration of the time /, there will result the 
following equations [80] : 

(m + m x + ;/z 2 -j- etc.) a = md-\- m x d x -f- m^ d 7 -j- etc., 

and (in -J- m x -\- m^ -f- etc.) x = m{d-\-vt) -\- m x (d. -f- v x t) -f- etc. 

Whence, by subtracting the first from the second, 

(m -j- m x -f- m % -f- etc.) (x — a) — (m v -\- m x v x -j- m^ v^ -f- etc.) /, 

(m v 4- m. v. 4- m„ v„ 4- etc.) 
or (x - a) = v , , l , ','T x t\ 

v ' {m 4- m x -f- m^ -f- etc.) 

and since the coefficient of / is constant, it follows that what- 
ever be the velocities of the elements m, m x , m» etc., com- 
posing the system, the distance (x — a) through which the 
centre of gravity moves is directly proportional to the time /, 
and therefore its motion will be uniform. 

It should also be observed that in the equations above the 
values of d, d v d„ etc., and v, v x , v v etc., will be either positive 



DYNAMICS. 255 

or negative according to the directions in which they are esti- 
mated. 

If we suppose the numerator of the above fraction to be- 
come zero, we shall have 

mv -\- m x v 1 -f- m^ v % + etc. = 2 m v = o, 

and hence (x — a) = o ; 

or, the centre of gravity will remain stationary whenever the alge- 
braic sum of the momenta of the various masses composing the 
system becomes equal to zero. 

If we denote by V the velocity of the centre of gravity, we 
have 

x — a 2 m v 2 m v 



V = 



2 m M 



where M\% the mass of the whole system. We have, there- 
fore, 

M V = 2 m v. 

But the first member of this equation represents the actual 
momentum of the whole system supposed to be concentrated 
at its centre of gravity, and measures the total impulse neces- 
sary to impart the velocity V actually received ; and this ag- 
gregated impulse is seen to be equal to the algebraic sum of 
the various impulses indicated by the second member. Hence 
we conclude that the centre of gravity moves as if the whole 
mass of the system were concentrated in it, and subjected to the 
combined actioji of all the separate forces of the system, in direc- 
tions parallel to those they are supposed to take. 

In the above discussion it will be observed that the im- 
pressed velocities were all supposed to be parallel, but a little 
consideration will show that the proposition just enunciated 
will also hold whatever directions they may take ; for if we 
suppose each velocity, or the force due to it, to be resolved 
into three others parallel to three rectangular axes, forming 



256 ELEMENTS OF MECHANICS. 

three groups of parallel forces or velocities, the foregoing 
reasoning would be applicable to each group, and the centre 
of gravity would move as if the three groups of forces, or the 
resultant of each, were simultaneously applied directly to it, 
and therefore as if all the impulses of the system were directly 
so applied, whilst preserving their original directions. 

[203.] Let us now consider the case in which all the 
bodies of the system are supposed to be rigidly connected, 
as with the elementary parts of a solid mass. 

Under this supposition each elementary portion would re- 
ceive some definite velocity due to the action of the impressed 
forces. Let F, F v F %i etc., be the forces corresponding or due 
to these velocities. Then, under D'Alembert's principle [191], 
these would be the effective forces of the system resulting 
from the combined action of the impressed forces, and the 
internal forces arising from the reciprocal action of the parts. 
Let P, P 1? P„ etc., be the impressed forces, and /, f, f, etc., 
the internal forces, so that F shall be the resultant of P and /, 
F x of P x and f, etc. Then, as already seen in the article re- 
ferred to, the forces /, f, f„ etc., mutually destroy each other 
and are in equilibrium. They will therefore satisfy the con- 
ditions of equilibrium of forces acting at points in space as 
given in section [66]. 

But these equations of condition include the equations 
necessary for equilibrium when the forces concur at one 
point. Hence f,f x , f, etc., may be transferred parallel to 
themselves to the centre of gravity, where also they will be 
of no effect. 

The forces F, F x , F v etc., due to the actual velocities of 
the parts of the system are therefore such that they would 
impart the very same velocities to the parts if they were en- 
tirely free and unconnected with each other. 

But in this case, by the preceding discussion, they may 
also be transferred parallel to themselves to the centre of 
gravity. 

Since, therefore, the components of P, P v P 2 , etc., may act at 




DYNAMICS. 257 

the centre of gravity of the mass without change of effect as far 
as regards its motion, so also may the impressed forces themselves. 
[204.] When a mass free to move is subject to the influ- 
ence of any number of forces the resultant of which does not 
pass through the centre of gravity, the motion of this centre, 
as we have already seen, will be the same as if this resultant 
were applied to it directly. But a motion of rotation would 
also ensue in precisely the same manner as if through the 
centre of gravity a fixed axis of revolution perpendicular to 
the planes of rotation existed. To show this, let R (Fig. 117) 
represent the resultant force applied 
at the point A of the body A B, free 
to move, and let G be its centre of 
gravity. Draw A G, and produce it 
it to B, making GB = GA, and ap- 
ply at B two equal and opposite 

forces, each equal to %R, and in the same plane as the force R. 
These forces will mutually destroy each other and have no 
disturbing effect upon the system. The effect of R alone is 
therefore equivalent to R acting at A and the two opposite 
forces %R, each applied at B. As has been shown, the mo- 
tion of translation due to R would be the same as if R acted 
at G, or, which is the same thing, as if \ R acted at A and \R 
in the same direction at B. But this would leave still a force 
\R acting at A, and the force \R acting in an opposite direc- 
tion at B, constituting a static couple, causing rotation about 
the centre of gravity G. Hence the force R alone would cause 
rotation about an axis passing through the centre of gravity, 
just as the couple referred to. The total moment of this cou- 
ple would plainly be equal to 

RXAG. 

It follows, therefore : 

1 st. That the motion of translation would be the same as if 
the resultant of the impressed forces, or these forces themselves, 
were applied directly to the centre of gravity ; 

2d. That the mass would rotate as if about a fixed axis 
through this centre. 



258 



ELEMENTS OF MECHANICS. 



[205.] If i£f denote the mass of the body and Fthe velo- 
city of the progressive motion, due to R applied at A or G, 
then we have as the measure of the force R 



R = MV, or 



M 



Likewise for the angular velocity go due to the force R 
supposed to act at A, at a distance A G = r from the centre 
of motion, we have [187] 



MVr MVr Vr 



GO = 



mr. 



Mk" 



in which k [191] is the principal radius of gyration. 

The actual velocity of any point within the mass is obvi- 
ously compounded of its rotary and progressive velocities, as 
given by the equations above, which may be thus briefly 
stated : 

The progressive velocity of the centre of gravity is equal to 
the moving force divided by the mass ; and the angular velocity 
is equal to the moment of the force about an axis through the cen- 
tre of gravity, divided by the moment of inertia of the mass with 
respect to the same axis. 

[206.] As already stated, the actual motion of any point 
in the mass is dependent both upon its angular and progres- 
sive motion. 

Let £7/ (Fig. 118) be the direction of the progressive mo- 
tion, and P Q and P l Q 1 the directions of the rotary motions of 
P and P x at an equal distance r, from the centre of gravity, at 
any given instant of time. Then 
the actual velocity of P will be, 
for that instant, 



AH 



and of P v 



V+r.co; 
V — r, go. 




mvsx 



Fig. 118. 



But since any point on the 
line GP V or this line produced, and supposed if necessary to 



DYNAMICS. 259 

be rigidly connected with the body, moves with a rotary 
velocity increasing with its distance from G, whilst the pro- 
gressive velocity V remains constant, it follows that there will 
be some point on this line, as m for instance, with reference 
to which the rotary and progressive velocities will be exactly 
equal. Around this point, then, the body will for an instant 
tend to rotate in its initial motion. This point is called the 
centre of spontaneous rotation. 

But under this supposition, if r x denote the distance of m 
from the point G, we have 

V — r x qd = o, or r. = — ; 
or, substituting the value of go already found above, 

r> = —> 

in which r = A G, the distance of the point of application of 
the force R from the centre of gravity G. 

Hence the distance of m, the centre of spontaneous rotation 
from the point of impact A, will be 

k? 
A m = r + 

By reference to section [195] it will be seen that the dis- 
tance of the centre of spontaneous rotation from the point of per- 
cussion is equal to the distance from this latter point, considered as 
a point of suspension, to the centre of oscillation of the body. It 
should be observed also that this distance is independent of 
the magnitude of the applied force. 

[207.] If it be required to determine at what distance 
from the centre of gravity the impulse must be given to pro- 
duce the actual progressive and rotary motions of the body 
denoted by V and 00, we have, from the preceding section, 

Vr Poo 

«'=-#> or r = nr- 



200 ELEMENTS OF MECHANICS. 

If now we suppose the rotary velocity of a point in the 

body at a distance R from the centre of gravity to be v, we 

have for this point 

v 
Rao = v and go = -5-, 

which gives by substitution above 

r ~RV' 

Example. — Let it be required to find at what distance from 
the centre of the earth an impulse should be given to impart the 
actual motion of translation, and of rotation really existing. 

In applying the formula 

fry 

we have R the equatorial radius of the earth, Fthe velocity 

of the earth in its orbit, and v the velocity of a point on the 

equator. The earth performs a revolution on its axis in one 

sidereal day, and its progressive velocity describes its orbit 

in 366 sidereal days. The circumference of the orbit being 

580,447,000 miles, and its equatorial circumference 24,886 

miles, we have 

24886 . TT 580447000 

v — and V = 



Whence 



366 
v 1 



V - 63.7 

The earth being regarded as a sphere, we have also, sec- 
tion [190, Cor.] and [190,/], 

79 2mrS 4 MR' 2 „, 
* ~ M ~ M ~ $ 
Hence, by substitution, 

k 2 v 2 „ 1 R 

r= -—X-F7 = -RX 



R ~ V ~ 5 63.7 159.2 ' 

that is to say, the point of impulse must be about the -j-^ part 
of the equatorial radius from the centre. 



APPENDIX A. 



[208.] To find the radius of curvature of the ellipse at any 
point, and the value of the expression Rp*, p being the perpendicu- 
lar upon the tangent from one focus. 

Referring the ellipse to conjugate diameters (Fig. 119), 




Fig. 119. 

suppose the osculatory circle drawn at P to coincide with the 
ellipse from P to k an indefinitely small arc. 

Let R == radius of curvature. 
a v b x = semi-conjugate axes. 
r, r 1 = radii-vectores. 
y = ordinate to ellipse. 
y -\- n = ordinate to circle. 
Pw=p,. 

Since in the ellipse the squares of any two ordinates are 
to each other as the product of the segments into which they 
divide the diameter, we have 

f : b? :: (2^ — x)x : a?\ 



262 APPENDIX. 

or, since the line x is at the limit indefinitely small, 

, K 2^ x 

y 1 — —^2a.x — — - 

Likewise, from a well-known property of the circle, 

(y + ny=(2R-z)z; 
or, since at the limit n and z are both indefinitely small lines, 

/ = 2Rz. 
Hence, by equating the values of y> 

_ b?x z b? 

Rz — — — , or 



x Ra x 
But from similar triangles, z : x :: / a : a v Hence 



$ = £? or A = §, or *A" = W. 

But by a property of the ellipse the rectangle upon the semi- 
axes a and b is equivalent to the parallelogram upon the semi- 
conjugate axes a x and b x . Hence 

ab = b x p 2 and a* b* = b? p 2 *. 

By substitution, Rp£ = a* b" 2 . 

This value of Rp£ applies to the case in which the centre of force 
is at o, the centre of the ellipse, section [172]. 

Again, to find Rp\ in which p is the perpendicular from 
the focus /, upon the tangent. Let p x be the perpendicular 
from the focus f. It is evident that 

A = OS=— 2 

But similar triangles give 

A_*\ 
p ~ r> 



APPENDIX. 



263 



whence 



f+Px r + r i 



P 



2 a 

T 9 



ap 



Hence A = ' — an ^ 

This, by substitution, gives 
Ra 3 p 3 



or 



p: 



P+P i _ £/ 
2 r 

a 3 p 3 



= « 



or 



^/■ = 



r 3 £ 2 



/^^ ^tfto required in the formula, section [155], when the central 
force is at the focus of the ellipse. 

The radius of curvature at the extremities of the major 
and minor axes is easily found from the above by giving r and 
p their proper values in the formulas. 

TO FIND Rp 3 IN THE CASE OF THE HYPERBOLA. 

The method does not differ from that in the case of the 
ellipse. Referring to the diagram (Fig. 120), you may derive 



N 




Fig. 120. 



for the hyperbola also, just as in the preceding case of the 
ellipse, 

x Ra t 

- = -7-j 2 , also x : z : : a x : p % ; 



264 



APPENDIX. 



whence Rp 2 = b x \ or Rp* = b*p a * = a*b\ 
But in this case 

and since here also p 1 p x :: r : r lf 

p l — p r 1 — r 2 a 



Hence 



P 



Pi-P _ *P _ . . 3 _ <**£_ 

2 — r —Pv P% ~ r 3 f 



and by substitution, 
Ra a p 3 



r 3 b* 

c?b\ and Rp % = —- y 



a 



the same value as for the ellipse. 

TO FIND Rp 3 IN THE CASE OF THE PARABOLA. 

As the parabola is but the limit to the ellipse when a and b 
become infinite, as shown by the manner in which it is cut 




Fig. 121. 



from the cone, it is obvious that the value of Rp 3 for the ellipse 
would apply also to the parabola, and the ratio of a to P or - 



APPENDIX. . 265 

would still be equal to the semi-latus rectum s. It is pro- 
posed, however, to give an independent solution. Supposing 
the curve (Fig. 121) to be referred to a tangent line at P and 
P ^parallel to the axis of the curve, and denoting by a the 
angle the tangent makes with the axis of the curve, we have 
for its equation 

« 2S 

y = -^—* — x. 

y sin a 
But r sin a — p, the perpendicular upon the tangent ; or 

sin 2 a — --- 

r 

2 S V* 

Hence y = — =— x. 

J P 

Also, in the equicurve circle at the limit, 

O-f nf =f = (2R -z)z = 2Rz. 

sr 2 
Hence Rz = ~r x. 

P 

But similar triangles Pqm and Pfv give 

x r r 

- — -, or x = -z. 
z i) p 

Hence R = — r and Rp % — r*s, 

P 

in which s is the semi-latus rectum. 



